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

AP Physics

Unit 6 - Rotational Motion

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Mathematical

FRQ

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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.

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Dynamic Torque, Rotational Motion, Torque

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In a demonstration, a teacher holds the axle of a wheel that is spinning with constant angular speed. The teacher then releases the axle and the wheel begins to fall toward the ground. As the wheel falls, its angular speed remains constant. Which of the following correctly describes how the rotational kinetic energy $ K_{\text{rot}} $ of the wheel and the total kinetic energy $ K_{\text{tot}} $ of the wheel change, if at all, after the wheel is released but before it reaches the ground?

	$ K_{\text{rot}} $	$ K_{\text{tot}} $
A	Constant	Constant
B	Constant	Increasing
C	Increasing	Constant
D	Increasing	Increasing

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The diagram above shows a top view of a child of mass $M$ on a circular platform of mass $2M$ that is rotating counterclockwise. Assume the platform rotates without friction. Which of the following describes an action by the child that will increase the angular speed of the platform-child system and gives the correct reason why?

Angular Momentum, Rotational Inertia, Rotational Motion

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Two masses, $ m_y = 32 $ $ \text{kg} $ and $ m_z = 38 $ $ \text{kg} $, are connected by a rope that hangs over a pulley. The pulley is a uniform cylinder of radius $ R = 0.311 $ $ \text{m} $ and mass $ 3.1 $ $ \text{kg} $. Initially, $ m_y $ is on the ground and $ m_z $ rests $ 2.5 $ $ \text{m} $ above the ground.

Rotational Energy, Rotational Inertia, Rotational Kinematics, Rotational Motion, Torque

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Consider a uniform hoop of radius $ R $ and mass $ M $ rolling without slipping. Which is larger, its translational kinetic energy or its rotational kinetic energy? Hint: The moment of inertia of a uniform hoop is $I = M R^2$

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

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

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

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.

	\( K_{\text{rot}} \)	\( K_{\text{tot}} \)
A	Constant	Constant
B	Constant	Increasing
C	Increasing	Constant
D	Increasing	Increasing

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]

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Unit 6 - Rotational Motion

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Part (a): Moment of Inertia

Part (b): Pulley Rotation

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