Rotational Motion

------------- | ----------------------------- | --------------- | | Displacement | Angle | | |” date: 2026-04-14 tags:

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$$\omega = \frac{d\theta}{dt}, \qquad \alpha = \frac{d\omega}{dt}$$$$\omega = \omega_0 + \alpha t$$ $$\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2$$ $$\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$$

For circular motion, the total acceleration has two components:

$$A_t = r\alpha \quad \mathrm{(tangential, changes speed)}$$ $$A_c = r\omega^2 = \frac{v^2}{r} \quad \mathrm{(centripetal, changes direction)}$$ $$A_{\mathrm{total}} = \sqrt{a_t^2 + a_c^2}$$

Every linear kinematic equation has a direct rotational analogue. Replace $x$ with $\theta$, $v$ with $\omega$ And $a$ with $\alpha$. This is not a coincidence: it reflects the fact that rotation Is a one-dimensional motion in the angular coordinate. The mathematics is identical.

Rotational motion is the mirror image of linear motion, every linear concept has a rotational analogue. Replace $x$ with $\theta$, $v$ with $\omega$, $a$ with $\alpha$, $m$ with $I$, and $F$ with $\tau$. The equations are identical in form. This isn’t a coincidence, it reflects the fact that rotation is one-dimensional motion in the angular coordinate.

Moment of inertia intuition: Just as mass resists linear acceleration, moment of inertia ($I$) resists angular acceleration. But unlike mass, $I$ depends on how the mass is distributed relative to the axis. A hollow cylinder has more $I$ than a solid one of the same mass because its mass is farther from the axis. This is why a figure skater spins faster when pulling in her arms, she reduces $I$, and angular momentum conservation forces $\omega$ to increase.

Rolling intuition: When a ball rolls without slipping, its kinetic energy splits between translation ($\frac{1}{2}mv^2$) and rotation ($\frac{1}{2}I\omega^2$). Objects with more mass near the rim (hollow cylinder) put more energy into rotation and less into translation, so they roll slower down an incline than solid objects.

Angular momentum conservation: When no external torque acts, $L = I\omega$ is constant. This is why gyroscopes precess, why planets move faster when closer to the Sun (Kepler’s second law), and why neutron stars spin incredibly fast.

1. Using the wrong moment of inertia. Always check which axis the object rotates about. Use the parallel axis theorem when the axis does not pass through the center of mass.
2. Confusing torque and force. Torque depends on both the force and the moment arm (distance from the axis).
3. Forgetting that angular momentum is a vector. Direction matters and is determined by the right-hand rule.
4. Incorrectly applying the rolling condition. Rolling without slipping means $v = R\omega$ Not $v = R\alpha$.
5. Not accounting for rotational KE in energy problems. Rolling objects have both translational and rotational kinetic energy.
6. Sign errors with torque. Be consistent with the sign convention (CCW positive, CW negative).
7. Assuming angular momentum is conserved when external torques act. Conservation requires zero net external torque.
8. Confusing moment of inertia with mass. Mass is a scalar; moment of inertia depends on the axis.
9. Using the wrong radius in $v = R\omega$. $R$ is the radius of the rolling object, not the radius of the incline or the path.

1. A disk of mass $8.0 \mathrm{ kg}$ and radius $0.5 \mathrm{ m}$ is mounted on a frictionless axle. A block of mass $2.0 \mathrm{ kg}$ is attached to a string wrapped around the disk. Find the acceleration of the block.

2. A solid cylinder and a hollow cylinder of the same mass and radius are released from rest at the top of an incline. Which reaches the bottom first? Prove your answer.

3. A uniform rod of length $L$ and mass $M$ is pivoted at one end and held horizontally, then released. Find its angular speed as it passes through the vertical position.

4. A merry-go-round of radius $3.0 \mathrm{ m}$ and moment of inertia $600 \mathrm{ kg \cdot \mathrm{m^2}}$ is rotating at $0.50 \mathrm{ rad/s}$. A child of mass $30 \mathrm{ kg}$ jumps on at the edge. Find the new angular speed.

5. A $4.0 \mathrm{ m}$ uniform beam of mass $50 \mathrm{ kg}$ is supported by two vertical cables, one at each end. A $200 \mathrm{ kg}$ mass hangs $1.0 \mathrm{ m}$ from the left end. Find the tension in each cable.

6. Calculate the moment of inertia of a uniform solid sphere about its diameter by integration.

7. A $0.50 \mathrm{ kg}$ ball of radius $0.05 \mathrm{ m}$ rolls without slipping down a $30^\circ$ incline from a height of $2.0 \mathrm{ m}$. Find the translational and rotational speeds at the bottom.

8. Derive the parallel axis theorem from the definition of moment of inertia.

9. A solid sphere rolls without slipping up a $20^\circ$ incline. If its initial speed is $5.0 \mathrm{ m/s}$ How far up the incline does it travel before stopping and rolling back?

10. A flywheel of moment of inertia $50 \mathrm{ kg \cdot \mathrm{m^2}}$ rotating at $300 \mathrm{ rpm}$ is brought to rest by a constant frictional torque of $10 \mathrm{ N \cdot \mathrm{m}}$ in $785 \mathrm{ s}$. Verify this using the rotational impulse equation, and calculate the angle through which the flywheel rotates while stopping.

Divide a solid sphere of mass $M$ and radius $R$ into thin disks of radius $r$ and thickness $dz$ at height $z$ from the centre.

$$r^2 = R^2 - z^2$$

The volume of each disk is $dV = \pi r^2 dz = \pi(R^2 - z^2)dz$.

$$dm = \frac{M}{\frac{4}{3}\pi R^3} \cdot \pi(R^2 - z^2)dz = \frac{3M}{4R^3}(R^2 - z^2)dz$$ $$I = \int_{-R}^{R} \frac{1}{2}r^2\, dm = \int_{-R}^{R} \frac{1}{2}(R^2 - z^2) \cdot \frac{3M}{4R^3}(R^2 - z^2)dz$$ $$= \frac{3M}{8R^3}\int_{-R}^{R} (R^2 - z^2)^2 dz = \frac{3M}{8R^3}\int_{-R}^{R} (R^4 - 2R^2z^2 + z^4)dz$$ $$= \frac{3M}{8R^3}\left[R^4 \cdot 2R - 2R^2 \cdot \frac{2R^3}{3} + \frac{2R^5}{5}\right] = \frac{3M}{8R^3}\left[2R^5 - \frac{4R^5}{3} + \frac{2R^5}{5}\right]$$ $$= \frac{3M}{8R^3} \cdot \frac{2R^5}{15}(15 - 10 + 3) = \frac{3M}{8R^3} \cdot \frac{16R^5}{15} = \frac{2MR^2}{5}$$

A thin uniform rod of mass $M$ and length $L$ Pivoted at one end.

$$I = \int_0^L x^2 \frac{M}{L}\, dx = \frac{M}{L}\left[\frac{x^3}{3}\right]_0^L = \frac{ML^2}{3}$$

Using the parallel axis theorem: $I_{\mathrm{end}} = I_{\mathrm{cm}} + Md^2 = \frac{ML^2}{12} + M\left(\frac{L}{2}\right)^2 = \frac{ML^2}{12} + \frac{ML^2}{4} = \frac{ML^2}{3}$. Confirmed.

A solid sphere, a solid cylinder, a hollow sphere, and a hollow cylinder, all of mass $M$ and radius $R$ Are released from rest at the top of an incline of height $h$. Rank them by their speed at the Bottom.

Ranking (fastest to slowest): solid sphere, solid cylinder, hollow sphere, hollow cylinder. Objects with more mass concentrated near the rim (larger $I$) have more rotational KE and less translational KE, so they move more slowly.

A solid sphere rolls without slipping up a $20^{\circ}$ incline with initial speed $5 \mathrm{ m/s}$. How far up does it travel?

$$Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}Mv^2 + \frac{1}{2}\cdot\frac{2}{5}MR^2\cdot\frac{v^2}{R^2} = \frac{7}{10}Mv^2$$ $$h = \frac{7v^2}{10g} = \frac{7 \times 25}{10 \times 9.8} = \frac{175}{98} = 1.786 \mathrm{ m}$$ $$d = \frac{h}{\sin 20^{\circ}} = \frac{1.786}{0.342} = 5.22 \mathrm{ m}$$

A $0.05 \mathrm{ kg}$ ball moving at $8 \mathrm{ m/s}$ strikes a rod of mass $2 \mathrm{ kg}$ and length $1 \mathrm{ m}$ that is free to rotate about one end. The ball strikes the rod at its free End, perpendicular to the rod. Find the angular velocity of the rod after the collision.

Conservation of angular momentum about the pivot (the external forces at the pivot produce no Torque about the pivot):

$$L_{\mathrm{initial}} = L_{\mathrm{final}}$$ $$mv \cdot L = I_{\mathrm{rod}\omega} + mv' \cdot L$$

Since the ball sticks to the end of the rod (perfectly inelastic collision):

$$mvr = (I_{\mathrm{rod}} + mr^2)\omega = \left(\frac{1}{3}Mr^2 + mr^2\right)\omega$$ $$\omega = \frac{mvr}{r^2\left(\frac{M}{3} + m\right)} = \frac{mv}{r\left(\frac{M}{3} + m\right)} = \frac{0.05 \times 8}{1 \times \left(\frac{2}{3} + 0.05\right)}$$ $$= \frac{0.4}{0.717} = 0.558 \mathrm{ rad/s}$$

A merry-go-round of radius $3 \mathrm{ m}$ and moment of inertia $600 \mathrm{ kg\cdot\mathrm{m^2}}$ Rotates at $0.5 \mathrm{ rad/s}$. A $60 \mathrm{ kg}$ person standing at the edge walks to the Centre. Find the new angular speed.

Initially: $I_i = 600 + 60 \times 3^2 = 600 + 540 = 1140 \mathrm{ kg\cdot\mathrm{m^2}}$.

Finally: $I_f = 600 + 60 \times 0^2 = 600 \mathrm{ kg\cdot\mathrm{m^2}}$.

$$I_i\omega_i = I_f\omega_f$$ $$1140 \times 0.5 = 600 \times \omega_f$$ $$\omega_f = \frac{570}{600} = 0.95 \mathrm{ rad/s}$$

The angular speed nearly doubles. This is the same principle behind figure skaters pulling in their arms to spin faster.

A uniform ladder of mass $15 \mathrm{ kg}$ and length $4 \mathrm{ m}$ leans against a smooth (frictionless) wall at $65^{\circ}$ to the horizontal. The floor is rough with $\mu_s = 0.4$. Will the ladder slip?

Take torques about the base of the ladder (eliminates the friction and normal force at the base):

Forces on the ladder:

• Weight $Mg$ at the centre ($2 \mathrm{ m}$ from base)
• Normal force from wall $N_w$ at the top ($4 \mathrm{ m}$ from base, horizontal)
• Normal force from floor $N_f$ at the base (vertical)
• Friction $f$ at the base (horizontal, towards wall)

Torque equation (about base):

Clockwise (positive): $Mg \times 2\cos 65^{\circ} = 15 \times 9.8 \times 2 \times 0.4226 = 124.2 \mathrm{ N\cdot\mathrm{m}}$

Anticlockwise (negative): $N_w \times 4\sin 65^{\circ} = N_w \times 3.625$

$$N_w \times 3.625 = 124.2 \implies N_w = 34.3 \mathrm{ N}$$

Horizontal equilibrium: $f = N_w = 34.3 \mathrm{ N}$

Vertical equilibrium: $N_f = Mg = 147 \mathrm{ N}$

Check slipping: $f_{\max} = \mu_s N_f = 0.4 \times 147 = 58.8 \mathrm{ N}$

Since $f = 34.3 \mathrm{ N \lt 58.8 \mathrm{ N}} = f_{\max}$ The ladder does not slip.

1. A uniform rod of mass $4 \mathrm{ kg}$ and length $2 \mathrm{ m}$ is pivoted at one end and held horizontally. It is released from rest. Find (a) the initial angular acceleration, (b) the angular speed as it passes through the vertical, and (c) the angular speed when it makes an angle of $45^{\circ}$ with the vertical.

2. A solid cylinder of mass $10 \mathrm{ kg}$ and radius $0.2 \mathrm{ m}$ rolls without slipping down an incline of $30^{\circ}$ from a height of $2 \mathrm{ m}$. Find the linear speed at the bottom and the time taken.

3. A disk of moment of inertia $0.5 \mathrm{ kg\cdot\mathrm{m^2}}$ rotating at $20 \mathrm{ rad/s}$ has a braking torque of $0.2 \mathrm{ N\cdot\mathrm{m}}$ applied. Find (a) the angular deceleration, (b) the time to stop, and (c) the number of revolutions made while stopping.

4. A $3 \mathrm{ m}$ uniform beam of mass $40 \mathrm{ kg}$ is hinged at a wall and supported by a cable at $30^{\circ}$ to the beam at the far end. A $100 \mathrm{ kg}$ mass hangs $1 \mathrm{ m}$ from the hinge. Find the tension in the cable and the force at the hinge.

5. Calculate the moment of inertia of a hollow cylinder (thin-walled) of mass $M$ and radius $R$ about its central axis by integration. Verify the result using the parallel axis theorem if necessary.

A spinning bicycle wheel of mass $2 \mathrm{ kg}$ and radius $0.35 \mathrm{ m}$ is supported on one end of its axle. The wheel spins at $50 \mathrm{ rad/s}$ and the axle is horizontal. The distance from the support to the wheel centre is $0.15 \mathrm{ m}$. Calculate the precession angular Velocity.

Step 1: Moment of inertia of the wheel (thin ring approximation)

$$I = MR^2 = 2 \times 0.35^2 = 0.245 \mathrm{ kg\cdot\mathrm{m^2}}$$

Step 2: Angular momentum of the wheel

$$L = I\omega = 0.245 \times 50 = 12.25 \mathrm{ kg\cdot\mathrm{m^2/\mathrm{s}}}$$

Step 3: Torque due to gravity about the support

$$\tau = Mg \times d = 2 \times 9.8 \times 0.15 = 2.94 \mathrm{ N\cdot\mathrm{m}}$$

Step 4: Precession angular velocity

$$\omega_p = \frac{\tau}{L} = \frac{2.94}{12.25} = 0.240 \mathrm{ rad/s}$$

The wheel precesses at $0.240 \mathrm{ rad/s}$ Completing one revolution in $T = 2\pi/\omega_p = 26.2 \mathrm{ s}$.

A uniform rod of mass $3 \mathrm{ kg}$ and length $2 \mathrm{ m}$ has a point mass of $2 \mathrm{ kg}$ attached at one end. Calculate the moment of inertia about (a) the end with the Point mass, and (b) the centre of the rod.

Part (a): About the end with the point mass

Rod about its end: $I_{\mathrm{rod}} = \frac{1}{3}ML^2 = \frac{1}{3} \times 3 \times 4 = 4 \mathrm{ kg\cdot\mathrm{m^2}}$

Point mass at the pivot: $I_{\mathrm{pm}} = mr^2 = 2 \times 0^2 = 0 \mathrm{ kg\cdot\mathrm{m^2}}$

$$I_a = 4 + 0 = 4 \mathrm{ kg\cdot\mathrm{m^2}}$$

Part (b): About the centre of the rod

Rod about its centre: $I_{\mathrm{rod}} = \frac{1}{12}ML^2 = \frac{1}{12} \times 3 \times 4 = 1 \mathrm{ kg\cdot\mathrm{m^2}}$

Point mass is $1 \mathrm{ m}$ from the centre: $I_{\mathrm{pm}} = 2 \times 1^2 = 2 \mathrm{ kg\cdot\mathrm{m^2}}$

$$I_b = 1 + 2 = 3 \mathrm{ kg\cdot\mathrm{m^2}}$$

Verify with parallel axis theorem (part a from part b):

$$I_a = I_b + M_{\mathrm{total}d^2} = 3 + 5 \times 1^2 = 8 \mathrm{ kg\cdot\mathrm{m^2}}$$

Wait — this gives $8$ Not $4$. Let me recheck. The parallel axis theorem requires the total mass To be at the centre of mass of the entire system, not just the rod.

Centre of mass from the pivot (end with point mass):

$$x_{\mathrm{cm}} = \frac{3 \times 1 + 2 \times 0}{5} = 0.6 \mathrm{ m from pivot}$$

Now using the parallel axis theorem from the centre of mass:

$$I_{\mathrm{cm}} = I_b = 3 \mathrm{ kg\cdot\mathrm{m^2 \quad \mathrm{(already calculated)}}}$$ $$I_a = I_{\mathrm{cm}} + M_{\mathrm{total} \times x_{\mathrm{cm}^2}} = 3 + 5 \times 0.36 = 3 + 1.8 = 4.8 \mathrm{ kg\cdot\mathrm{m^2}}$$

This still does not match. Let me recalculate $I_b$ more carefully. The point mass is at the end Of the rod, which is $1 \mathrm{ m}$ from the centre of the rod. So $I_b = 1 + 2 \times 1^2 = 3$ is correct.

And $I_a$ directly: rod about its far end (away from point mass): use parallel axis from centre, $I_{\mathrm{rod, end}} = \frac{1}{12}(3)(4) + 3(1)^2 = 1 + 3 = 4$. Point mass at distance $2 \mathrm{ m}$: $I_{\mathrm{pm}} = 2 \times 4 = 8$. Wait — the point mass is at the pivot, so $r = 0$ Giving $I_a = 4 + 0 = 4$.

The parallel axis check failed because I was not careful. The correct check: $I_{\mathrm{cm}} = 3$ about the centre of mass at $0.6 \mathrm{ m}$ from pivot, so $I_a = 3 + 5(0.6)^2 = 3 + 1.8 = 4.8$. but direct calculation gives $4$.

Let me recheck $I_b$: the point mass is at one end of the rod, which is $1 \mathrm{ m}$ from the rod’s centre. $I_{\mathrm{pm}} = 2 \times 1^2 = 2$. $I_{\mathrm{rod, centre}} = 1$. $I_b = 3$. Correct.

The discrepancy means $I_{\mathrm{cm} \ne I_b}$. The centre of mass of the system is at $0.6 \mathrm{ m}$ from the pivot (not at the centre of the rod). So $I_{\mathrm{cm}}$ should be Calculated about the system’s centre of mass, not the rod’s centre. This is a subtle but important Point.

A yo-yo consists of two disks of total mass $0.1 \mathrm{ kg}$ and radius $3 \mathrm{ cm}$ With an Axle of radius $0.5 \mathrm{ cm}$. The string unwinds from the axle. Find the acceleration of the Yo-yo as it falls and the tension in the string.

Step 1: Equations of motion

Translation: $mg - T = ma$

Rotation: $TR = I\alpha$ Where $R$ is the axle radius and $a = R\alpha$.

$$T \times 0.005 = I \times \frac{a}{0.005}$$

For the yo-yo (solid disk): $I = \frac{1}{2}MR_{\mathrm{disk}^2} = \frac{1}{2} \times 0.1 \times 0.03^2 = 4.5 \times 10^{-5} \mathrm{ kg\cdot\mathrm{m^2}}$

$$T = \frac{Ia}{R^2} = \frac{4.5 \times 10^{-5} \times a}{2.5 \times 10^{-5}} = 1.8a$$

Step 2: Substitute into translational equation

$$mg - 1.8a = ma$$ $$a = \frac{mg}{m + 1.8} = \frac{0.1 \times 9.8}{0.1 + 1.8} = \frac{0.98}{1.9} = 0.516 \mathrm{ m/s^2}$$ $$T = 1.8 \times 0.516 = 0.929 \mathrm{ N}$$

In problems involving wheels, pulleys, or drums, the radius used in $\tau = Fr$ must be the radius at which the force is applied, which may differ from the overall radius. For example, a force Applied at a string wound around an axle uses the axle radius, not the wheel radius.

For rolling without slipping, $v = r\omega$ relates the centre-of-mass velocity to the angular Velocity about the centre of mass. A point on the rim has a velocity of $2v$ at the top and $0$ at the contact point (instantaneously at rest).

Moment of inertia has units of $\mathrm{kg\cdot\mathrm{m^2}}$. A common error is to use centimetres instead of metres when calculating $I = mr^2$ Giving answers that are off by a factor of $10^4$. always convert to SI units before calculating.

1. A solid sphere of mass $4 \mathrm{ kg}$ and radius $0.1 \mathrm{ m}$ rolls without slipping up a $20^\circ$ incline at $5 \mathrm{ m/s}$. Calculate how far up the incline it travels before stopping and rolling back.

2. A figure skater with arms extended has moment of inertia $4.5 \mathrm{ kg\cdot\mathrm{m^2}}$ and rotates at $2 \mathrm{ rad/s}$. She pulls her arms in, reducing her moment of inertia to $1.5 \mathrm{ kg\cdot\mathrm{m^2}}$. Calculate her new angular velocity and the ratio of final to initial rotational KE.

3. A uniform rod of mass $8 \mathrm{ kg}$ and length $3 \mathrm{ m}$ is hinged at one end and held horizontally. It is released from rest. Calculate the angular acceleration just after release, the angular velocity as it passes through the vertical, and the force at the hinge at that instant.

4. A $500 \mathrm{ g}$ ball of radius $5 \mathrm{ cm}$ rolls without slipping along a horizontal surface at $4 \mathrm{ m/s}$. It encounters a ramp of height $0.5 \mathrm{ m}$. Does it reach the top? If so, what is its speed at the top?

5. Two flywheels ($I_1 = 2 \mathrm{ kg\cdot\mathrm{m^2}}$ spinning at $300 \mathrm{ rpm}$ $I_2 = 5 \mathrm{ kg\cdot\mathrm{m^2}}$ at rest) are coupled together. Calculate the final angular velocity and the energy lost in the coupling process.

Question 1: Rolling without slipping down an incline

A solid sphere of mass $2 \mathrm{ kg}$ and radius $0.1 \mathrm{ m}$ rolls without slipping down a $30^\circ$ incline from rest. Calculate (a) the acceleration of the centre of mass, (b) the angular Acceleration, and (c) the speed after rolling $3 \mathrm{ m}$ along the incline.

Answer

For a solid sphere, $I = \frac{2}{5}mr^2$.

Acceleration: $a = \frac{g\sin\theta}{1 + I/(mr^2)} = \frac{g\sin\theta}{1 + 2/5} = \frac{9.8 \times 0.5}{1.4} = 3.5 \mathrm{ m/s^2}$.

Angular acceleration: $\alpha = a/r = 3.5/0.1 = 35 \mathrm{ rad/s^2}$.

Speed: $v^2 = 2ad = 2 \times 3.5 \times 3 = 21$, $v = 4.58 \mathrm{ m/s}$.

Question 2: Conservation of angular momentum

A figure skater with arms extended has a moment of inertia of $4.5 \mathrm{ kg\cdot m^2}$ and Rotates at $2 \mathrm{ rad/s}$. When she pulls her arms in, her moment of inertia decreases to $1.5 \mathrm{ kg\cdot m^2}$. Calculate her new angular velocity and the ratio of her new kinetic Energy to the original.

Answer

Conservation of angular momentum: $L = I_1\omega_1 = I_2\omega_2$.

$\omega_2 = \frac{I_1\omega_1}{I_2} = \frac{4.5 \times 2}{1.5} = 6 \mathrm{ rad/s}$.

KE ratio: $\frac{KE_2}{KE_1} = \frac{\frac{1}{2}I_2\omega_2^2}{\frac{1}{2}I_1\omega_1^2} = \frac{1.5 \times 36}{4.5 \times 4} = \frac{54}{18} = 3$.

The kinetic energy triples. The extra energy comes from the work done by the skater in pulling her arms inward against the centrifugal tendency.

Question 3: Torque and equilibrium of a beam

A uniform beam of length $4 \mathrm{ m}$ and mass $20 \mathrm{ kg}$ is supported at its left end by A hinge and at a point $1 \mathrm{ m}$ from the right end by a cable making $30^\circ$ with the horizontal. A $50 \mathrm{ kg}$ mass hangs from the right end. Find the tension in the cable and the force exerted by the hinge.

Answer

Taking torques about the hinge (left end):

Forces and distances from hinge:

• Weight of beam: $20g$ at $2 \mathrm{ m}$.
• Hanging mass: $50g$ at $4 \mathrm{ m}$.
• Cable tension $T$ at $3 \mathrm{ m}$ from hinge, at $30^\circ$ above horizontal. Vertical component = $T\sin(30^\circ) = T/2$.

Clockwise torques: $20g \times 2 + 50g \times 4 = 40g + 200g = 240g = 2352 \mathrm{ N\cdot m}$.

Anticlockwise torque: $T\sin(30^\circ) \times 3 = 1.5T$.

$1.5T = 2352$ So $T = 1568 \mathrm{ N}$.

Hinge force: Horizontal = $T\cos(30^\circ) = 1568 \times 0.866 = 1358 \mathrm{ N}$ (outward). Vertical = $70g - T\sin(30^\circ) = 686 - 784 = -98 \mathrm{ N}$ (downward).

Question 4: Physical pendulum

A uniform rod of length $1.0 \mathrm{ m}$ and mass $2 \mathrm{ kg}$ is pivoted at one end and swings as a physical pendulum. Calculate the period of small oscillations.

Answer

Moment of inertia about the end: $I = \frac{1}{3}mL^2 = \frac{1}{3}(2)(1)^2 = 0.667 \mathrm{ kg\cdot m^2}$.

Distance from pivot to centre of mass: $d = L/2 = 0.5 \mathrm{ m}$.

Period: $T = 2\pi\sqrt{\frac{I}{mgd}} = 2\pi\sqrt{\frac{0.667}{2 \times 9.8 \times 0.5}} = 2\pi\sqrt{\frac{0.667}{9.8}} = 2\pi\sqrt{0.0681} = 2\pi \times 0.261 = 1.64 \mathrm{ s}$.

Question 5: Angular momentum of a system

A disk of mass $3 \mathrm{ kg}$ and radius $0.4 \mathrm{ m}$ rotates at $10 \mathrm{ rad/s}$. A $1 \mathrm{ kg}$ lump of clay is dropped onto the disk at a distance of $0.3 \mathrm{ m}$ from the Centre and sticks. What is the new angular velocity?

Answer

Initial angular momentum: $L = I_{\mathrm{disk}\omega} = \frac{1}{2}(3)(0.4)^2 \times 10 = 0.24 \times 10 = 2.4 \mathrm{ kg\cdot m^2/s}$.

Final moment of inertia: $I_f = I_{\mathrm{disk}} + mr^2 = 0.24 + 1(0.3)^2 = 0.24 + 0.09 = 0.33 \mathrm{ kg\cdot m^2}$.

Conservation: L_f = L_i$$I_f\omega_f = 2.4$$\omega_f = 2.4/0.33 = 7.27 \mathrm{ rad/s}.

• Newton’s Laws — Rotational dynamics extends Newton’s second law to torque and angular acceleration.
• Work, Energy, and Power — Rotational kinetic energy and rotational work parallel the translational concepts.
• Momentum and Impulse — Angular momentum conservation connects rotational motion to the conservation principles.
• AP Calculus, Integrals: Moment of inertia and rotational work are computed via integrals over continuous mass distributions, applying integration to rotational systems.