Two masses, $ m_1 $ and $ m_2 $, are suspended on either side of a pulley with a radius $ R $, as shown. The heavier mass, $ m_2 $, is initially held at rest above the ground by a distance $ h $ before being released. An student measures that it takes an amount of time $ t $ for the heavier mass to hit the ground after being released.(a) The pulley is an odd shape, so the student would like to measure its moment of inertia. Derive an algebraic expression for the moment of inertia of the pulley in terms of only quantities measured or known by the student ($ m_1, m_2, R, h, t $ and physical constants). , (b) How many radians has the pulley rotated from the time it was released to the time $ m_2 $ hit the ground? Express your answer in terms of $ m_1, m_2, R, h, t$ and physical constants. :

$\frac{R^{2}t^{2}}{2h}\Bigl[(m_{2}-m_{1})g\Bigr] – R^{2}(m_{2}+m_{1})$ $\frac{h}{R}$

Two masses, $ m_1 $ and $ m_2 $, are suspended on either side of a pulley with a radius $ R $, as shown. The heavier mass, $ m_2 $, is initially held at rest above the ground by a distance $ h $ before being released. An experimenting elephant measures that it takes an amount of time $ t $ for the heavier mass to hit the ground after being released.\n(a) The pulley is an odd shape, and the elephant would like to measure its moment of inertia. Derive an algebraic expression for the moment of inertia of the pulley in terms of only quantities measured or known by the elephant ($ m_1, m_2, R, h, t $ and physical constants).\n(b) How many radians has the pulley rotated from the time it was released to the time $ m_2 $ hit the ground? Express your answer in terms of $ m_1, m_2, R, h $ and physical constants. • Nerd Notes

New Tool FRQ Atlas - Find, Solve, and Grade Any FRQ In Seconds.

AP Physics

Unit 6 - Rotational Motion

Advanced

Mathematical

FRQ

Answer Feedback from Phy

Verfied Answer

Verfied Explanation 0 likes

Part (a): Moment of Inertia

Step	Derivation/Formula	Reasoning
1	$$ T_{1} – m_{1}g = m_{1}a $$	This is Newton’s second law for mass $m_{1}$ moving upward.
2	$$ m_{2}g – T_{2} = m_{2}a $$	This is Newton’s second law for mass $m_{2}$ moving downward.
3	$$ T_{1} = m_{1}g + m_{1}a \quad \text{and} \quad T_{2} = m_{2}g – m_{2}a $$	Rearrange the equations to solve for the tensions in the string.
4	$$ (T_{2} – T_{1})R = I\left(\frac{a}{R}\right) $$	This relates the net torque on the pulley to its moment of inertia $I$ using the no‐slip condition $\alpha = \frac{a}{R}$.
5	$$ T_{2} – T_{1} = \frac{I\,a}{R^{2}} $$	Simplify the torque equation by dividing both sides by $R$.
6	$$ (m_{2}g – m_{2}a) – (m_{1}g + m_{1}a) = $$$$(m_{2}-m_{1})g – (m_{2}+m_{1})a =$$$$\frac{I\,a}{R^{2}} $$	Substitute the expressions for $T_{1}$ and $T_{2}$ into the torque equation.
7	$$ I = \frac{R^{2}}{a}\Bigl[(m_{2}-m_{1})g – (m_{2}+m_{1})a\Bigr] $$	Rearrange the equation to solve for the moment of inertia $I$.
8	$$ h = \frac{1}{2}at^{2} $$	Use the kinematics relation for the heavy mass $m_{2}$ falling a distance $h$ from rest.
9	$$ a = \frac{2h}{t^{2}} $$	Solve for the acceleration $a$ from the kinematics equation.
10	$$ I = \frac{R^{2}}{\frac{2h}{t^{2}}}\Bigl[(m_{2}-m_{1})g – (m_{2}+m_{1})\frac{2h}{t^{2}}\Bigr] $$	Substitute $a = \frac{2h}{t^{2}}$ into the expression for $I$.
11	$$ I = \frac{R^{2}t^{2}}{2h}\Bigl[(m_{2}-m_{1})g\Bigr] – R^{2}(m_{2}+m_{1}) $$	Simplify the expression to obtain $I$ solely in terms of $m_{1}, m_{2}, R, h, t$ and $g$.
12	$$ \boxed{I = \frac{R^{2}t^{2}}{2h}\Bigl[(m_{2}-m_{1})g\Bigr] – R^{2}(m_{2}+m_{1})} $$	This is the final algebraic expression for the pulley’s moment of inertia.

Part (b): Pulley Rotation

Step	Derivation/Formula	Reasoning
1	$$ \Delta x = R\theta $$	This equation relates the linear displacement $\Delta x$ to the angular displacement $\theta$ of the pulley.
2	$$ h = R\theta $$	Since the heavy mass $m_{2}$ falls a distance $h$, the length of the unwound rope is $h$, which equals $R\theta$.
3	$$ \theta = \frac{h}{R} $$	Solve for the angular displacement $\theta$ of the pulley.
4	$$ \boxed{\theta = \frac{h}{R}} $$	This is the final expression for the total rotation of the pulley in radians.

Need Help? Ask Phy To Explain

Just ask: "Help me solve this problem."

Just Drag and Drop!

Quick Actions ?

Topics in this question

Dynamic Torque, Rotational Motion, Torque

Join 1-to-1 Elite Tutoring

See how Others Did on this question | Coming Soon

Related Questions

5 More similar questions to the one you just solved

Beginner

Conceptual

MCQ

A point P is at a distance $ R $ from the axis of rotation of a rigid body whose angular velocity and angular acceleration are $ \omega $ and $ \alpha $ respectively. The linear speed, centripetal acceleration, and tangential acceleration of the point can be expressed as:

	Linear speed	Centripetal acceleration	Tangential acceleration
$ (a) $	$ R\omega $	$ R\omega^{2} $	$ R\alpha $
$ (b) $	$ R\omega $	$ R\alpha $	$ R\omega^{2} $
$ (c) $	$ R\omega^{2} $	$ R\alpha $	$ R\omega $
$ (d) $	$ R\omega $	$ R\omega^{2} $	$ R\omega $
$ (e) $	$ R\omega^{2} $	$ R\alpha $	$ R\omega^{2} $

Centripetal Acceleration, Rotational Kinematics, Rotational Motion

Advanced

Mathematical

FRQ

A uniform ladder with mass $ m_2 $ and length $ L $ rests against a smooth wall. A do-it-yourself enthusiast of mass $ m_1 $ stands on the ladder a distance $ d $ from the bottom (measured along the ladder). The ladder makes an angle $ \theta $ with the ground. There is no friction between the wall and the ladder, but there is a frictional force of magnitude $ f $ between the floor and the ladder. $ N_1 $ is the magnitude of the normal force exerted by the wall on the ladder, and $ N_2 $ is the magnitude of the normal force exerted by the ground on the ladder. Throughout the problem, consider counterclockwise torques to be positive.

Rotational Motion, Torque

Intermediate

Mathematical

MCQ

Consider a solid uniform sphere of radius $R$ and mass $M$ rolling without slipping. Which form of its kinetic energy is larger, translational or rotational?

Rotational Energy, Rotational Motion

Beginner

Proportional Analysis

MCQ

A $ 4 $-$ \text{kg} $ ball and a $ 1 $-$ \text{kg} $ ball are positioned a distance $ L $ apart on a bar of negligible mass. How far from the $ 4 $-$ \text{kg} $ mass should the fulcrum be placed to balance the bar?

Static Torque, Torque

Intermediate

Conceptual

MCQ

The diagram above shows a top view of a child of mass $M$ on a circular platform of mass $5M$that is rotating counterclockwise. Assume the platform rotates without friction. Which of the following describes an action by the child that will result in an increase in the total angular momentum of the child-platform system?

Angular Momentum, Rotational Motion

$\frac{R^{2}t^{2}}{2h}\Bigl[(m_{2}-m_{1})g\Bigr] – R^{2}(m_{2}+m_{1})$
$\frac{h}{R}$

	Linear speed	Centripetal acceleration	Tangential acceleration
\( (a) \)	\( R\omega \)	\( R\omega^{2} \)	\( R\alpha \)
\( (b) \)	\( R\omega \)	\( R\alpha \)	\( R\omega^{2} \)
\( (c) \)	\( R\omega^{2} \)	\( R\alpha \)	\( R\omega \)
\( (d) \)	\( R\omega \)	\( R\omega^{2} \)	\( R\omega \)
\( (e) \)	\( R\omega^{2} \)	\( R\alpha \)	\( R\omega^{2} \)

Kinematics	Forces
\(\Delta x = v_i t + \frac{1}{2} at^2\)	\(F = ma\)
\(v = v_i + at\)	\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)	\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)	\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\)

Circular Motion	Energy
\(F_c = \frac{mv^2}{r}\)	\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)	\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)	\(KE_i + PE_i = KE_f + PE_f\)
	\(W = Fd \cos\theta\)

Momentum	Torque and Rotations
\(p = mv\)	\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)	\(I = \sum mr^2\)
\(p_i = p_f\)	\(L = I \cdot \omega\)

Simple Harmonic Motion	Fluids
\(F = -kx\)	\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)	\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)	\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)	\(F_b = \rho V g\)
\(a = -\omega^2 x\)	\(A_1v_1 = A_2v_2\)

Step	Derivation/Formula	Reasoning
1	$$ T_{1} – m_{1}g = m_{1}a $$	This is Newton’s second law for mass \(m_{1}\) moving upward.
2	$$ m_{2}g – T_{2} = m_{2}a $$	This is Newton’s second law for mass \(m_{2}\) moving downward.
3	$$ T_{1} = m_{1}g + m_{1}a \quad \text{and} \quad T_{2} = m_{2}g – m_{2}a $$	Rearrange the equations to solve for the tensions in the string.
4	$$ (T_{2} – T_{1})R = I\left(\frac{a}{R}\right) $$	This relates the net torque on the pulley to its moment of inertia \(I\) using the no‐slip condition \(\alpha = \frac{a}{R}\).
5	$$ T_{2} – T_{1} = \frac{I\,a}{R^{2}} $$	Simplify the torque equation by dividing both sides by \(R\).
6	$$ (m_{2}g – m_{2}a) – (m_{1}g + m_{1}a) = $$$$(m_{2}-m_{1})g – (m_{2}+m_{1})a =$$$$\frac{I\,a}{R^{2}} $$	Substitute the expressions for \(T_{1}\) and \(T_{2}\) into the torque equation.
7	$$ I = \frac{R^{2}}{a}\Bigl[(m_{2}-m_{1})g – (m_{2}+m_{1})a\Bigr] $$	Rearrange the equation to solve for the moment of inertia \(I\).
8	$$ h = \frac{1}{2}at^{2} $$	Use the kinematics relation for the heavy mass \(m_{2}\) falling a distance \(h\) from rest.
9	$$ a = \frac{2h}{t^{2}} $$	Solve for the acceleration \(a\) from the kinematics equation.
10	$$ I = \frac{R^{2}}{\frac{2h}{t^{2}}}\Bigl[(m_{2}-m_{1})g – (m_{2}+m_{1})\frac{2h}{t^{2}}\Bigr] $$	Substitute \(a = \frac{2h}{t^{2}}\) into the expression for \(I\).
11	$$ I = \frac{R^{2}t^{2}}{2h}\Bigl[(m_{2}-m_{1})g\Bigr] – R^{2}(m_{2}+m_{1}) $$	Simplify the expression to obtain \(I\) solely in terms of \(m_{1}, m_{2}, R, h, t\) and \(g\).
12	$$ \boxed{I = \frac{R^{2}t^{2}}{2h}\Bigl[(m_{2}-m_{1})g\Bigr] – R^{2}(m_{2}+m_{1})} $$	This is the final algebraic expression for the pulley’s moment of inertia.

Constant	Description
[katex]g[/katex]	Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]	Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]	Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]	Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]	Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]	Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]	Mass of the Sun

Variable	SI Unit
[katex]s[/katex] (Displacement)	[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)	[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)	[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)	[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)	[katex]\text{kilograms (kg)}[/katex]

Variable	Derived SI Unit
[katex]F[/katex] (Force)	[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)	[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)	[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)	[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)	[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)	[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)	[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)	[katex]\text{hertz (Hz)}[/katex]

Prefix	Symbol	Power of Ten
Pico-	p	[katex]10^{-12}[/katex]
Nano-	n	[katex]10^{-9}[/katex]
Micro-	µ	[katex]10^{-6}[/katex]
Milli-	m	[katex]10^{-3}[/katex]
Centi-	c	[katex]10^{-2}[/katex]
Deci-	d	[katex]10^{-1}[/katex]
(Base unit)	–	[katex]10^{0}[/katex]
Deca- or Deka-	da	[katex]10^{1}[/katex]
Hecto-	h	[katex]10^{2}[/katex]
Kilo-	k	[katex]10^{3}[/katex]
Mega-	M	[katex]10^{6}[/katex]
Giga-	G	[katex]10^{9}[/katex]
Tera-	T	[katex]10^{12}[/katex]

Everything you need to master Physics — for free.

Meet Phy. The world's most advanced AI helper.

1000+ Physics problems + AI help to supercharge learning.

8 AP Physics units condensed into 8 short review guides.

Proven tips, tricks, and hacks to help you score that coveted 5.

Learn Physics, from scratch, at your own pace.

Guaranteed performance boost in scores and understanding.

5 weeks for a 5. Expedited progress. Stellar Results.

100+ hours of guided instruction to simplify physics & boost grades

Other 1-to-1 and group programs to accelerate Physics mastery.

Self Paced. 100+ hours of lessons, worked examples, and targeted practice.

A custom-built platform to simplify and accelerate your learning journey.

New Tool FRQ Atlas - Find, Solve, and Grade Any FRQ In Seconds.

AP Physics

Unit 6 - Rotational Motion

Advanced

Mathematical

FRQ

You're a Pro Member

Supercharge UBQ

Part (a): Moment of Inertia

Part (b): Pulley Rotation

Need Help? Ask Phy To Explain

Topics in this question

Join 1-to-1 Elite Tutoring

See how Others Did on this question | Coming Soon

Related Questions

5 More similar questions to the one you just solved

Nerd Notes

Discover the world's best Physics resources

Continue with

Error Report

Sign In to View Your Questions

Share This Question

Phy Pro

$11.99

Physics is Hard, But It Does NOT Have to Be

📚 Predict Your AP Physics Exam Score

You can close this ad in 5 seconds.

Ads show frequently. Upgrade to Phy Pro to remove ads.

You can close this ad in 7 seconds.

Ads display every few minutes. Upgrade to Phy Pro to remove ads.

You can close this ad in 5 seconds.

Ads show frequently. Upgrade to Phy Pro to remove ads.

How to UBQ

1. UBQ Credits

2. Auto Track Stats

3. Point System

Phy helps customize your learning experience.

4. Question Explantions

Phy gives feedback.

4. Feedback

We use cookies to improve your experience. By continuing to browse on Nerd Notes, you accept the use of cookies as outlined in our privacy policy.