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Chapter 10: Problem 41

A hollow, thin-walled sphere of mass \(12.0 \mathrm{~kg}\) and diameter \(48.0 \mathrm{~cm}\) is rotating about an axle through its center. The angle (in radians) through which it turns as a function of time (in seconds) is given by \(\theta(t)=A t^{2}+B t^{4},\) where \(A\) has numerical value 1.50 and \(B\) has numerical value \(1.10 .\) (a) What are the units of the constants \(A\) and \(B ?\) (b) At the time \(3.00 \mathrm{~s}\), find (i) the angular momentum of the sphere and (ii) the net torque on the sphere.

Short Answer

Expert verified
The units for A and B are \(rad/s^{2}\) and \(rad/s^{4}\) respectively. The angular momentum of the sphere at 3.00 s is \(30.8397 kgm^{2}/s\) and the net torque on the sphere at this time is \(30.02 Nm\).

Step by step solution

01

Find the Units for A and B

In the equation \(\theta(t)= A t^{2}+ B t^{4}\), \(\theta(t)\) represents angular displacement measured in radians and \(t\) represents time in seconds. For \(A t^{2}\), A must have units that when multiplied with time squared (\(s^{2}\)) gives angle in radians. Hence, the units for \(A\) are \(rad/s^{2}\). Similarly, for \(B t^{4}\), to get radians as the result, B must have units \(rad/s^{4}\).
02

Compute the Sphere's Moment of Inertia

The moment of inertia \(I\) for a hollow sphere is given by \(I = \frac{2}{3}mr^{2}\), where \(m\) is the mass and \(r\) is the radius. Here, \(m = 12.0 kg\) and \(r = \frac{48.0 cm}{2} = 24.0 cm = 0.24 m\). Substituting the values, we get \(I = \frac{2}{3}(12.0 kg)(0.24 m)^{2} = 0.6912 kgm^{2}\) .
03

Calculate the Angular Velocity

The angular velocity \(\omega\) at any time \(t\) can be found by differentiating the equation for \(\theta(t)\) with respect to \(t\). So, \(\omega(t) = \frac{d\theta}{dt} = 2A t + 4B t^{3}\). Substituting \(A = 1.5 rad/s^{2}\), \(B = 1.1 rad/s^{4}\), and \(t = 3.0 s\), we find \(\omega = 2(1.5 rad/s^{2})*3.0s + 4(1.1 rad/s^{4})*(3.0s)^{3} = 9 rad/s + 35.64 rad/s = 44.64 rad/s\).
04

Compute the Angular Momentum

Angular momentum \(L\) is the product of the moment of inertia \(I\) and the angular velocity \(\omega\). Substituting \(I = 0.6912 kgm^{2}\) and \(\omega = 44.64 rad/s\), we get \(L = 0.6912 kgm^{2} * 44.64 rad/s = 30.8397 kgm^{2}/s\).
05

Calculate the Net Torque

The net torque \(\tau\) on an object is equal to the derivative of its angular momentum with respect to time. So, \(\tau = \frac{dL}{dt}\). Differentiating the equation for \(L(t)\) = I*2A t + 4B t^{3}, with \(I = 0.6912 kgm^{2}\), \(A = 1.5 rad/s^{2}\), and \(B = 1.1 rad/s^{4}\), and evaluating at \(t = 3.0 s\), we find \(\tau = 0.6912 kgm^{2} * (2*1.5 rad/s^{2} + 12*1.1 rad/s^{4}*3.0s^{2}) = 30.02 Nm \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Displacement
Imagine spinning a wheel: the angle it turns through is a measure of angular displacement. In physics, it鈥檚 the angle through which a point or line has been rotated in a specified sense about a specified axis. It is commonly measured in radians or degrees. In our exercise, a formula is given for angular displacement as a function of time, \( \theta(t) = At^2 + Bt^4 \), where time \( t \) is squared and to the fourth power, showing that the displacement doesn鈥檛 change uniformly but accelerates over time.

This is significant because it shows a non-linear relationship between time and the angular displacement of the sphere, meaning as time goes on, the angle through which the sphere has turned increases at a faster rate, a concept that is fundamental to understanding motions such as that of planets in orbits or wheels in machines.
Moment of Inertia
The resistance of an object to change its state of rotational motion or angular velocity is quantified by its moment of inertia. Think of it as the rotational equivalent to mass in linear motion. For a hollow sphere with mass \( m \) and radius \( r \) rotating around an axis through its center, the moment of inertia \( I \) is given by \( I = \frac{2}{3}mr^2 \).

The moment of inertia depends on the distribution of the object's mass relative to the axis of rotation: the further the mass is from the axis, the larger the moment of inertia. Hence in our example, using the given mass and diameter (converted to radius), we can calculate this sphere's resistance to changes in its rotational motion.
Angular Momentum
For any moving object, we can talk about its linear momentum. But when the object is spinning, it has angular momentum, denoted as \( L \). It takes into account not just the velocity (in this case, angular velocity) but also the mass distribution (moment of inertia). The angular momentum of an object can be calculated by \( L = I\omega \) where \( \omega \) is the angular velocity.

In our hollow sphere, once the moment of inertia is known and the angular velocity is calculated by differentiating the angular displacement with respect to time, the sphere鈥檚 angular momentum can be determined. It represents the 'quantity' of rotation of the sphere, not just the rate at which it spins.
Net Torque
Torque is the twisting force that causes rotation. The net torque is the sum of all torques acting on an object, which determines the object's angular acceleration. Newton's second law for rotation states that net torque equals the change in angular momentum over time, \( \tau = \frac{dL}{dt} \).

This relationship is used in our exercise to find the net torque on the sphere at a specific time. By differentiating the equation \( L(t) = I(2At + 4Bt^3) \) and substituting the known values, we calculate the net torque, which tells us how much the sphere's angular velocity is changing at that moment. It鈥檚 a crucial concept when analyzing how forces affect rotational motion, like in engines or turbines.

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Most popular questions from this chapter

A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equal to \(80.0 \mathrm{~N}\) is applied to the rim of the wheel. The wheel has radius \(0.120 \mathrm{~m}\) Starting from rest, the wheel has an angular speed of \(12.0 \mathrm{rev} / \mathrm{s}\) after \(2.00 \mathrm{~s}\). What is the moment of inertia of the wheel?

A solid ball is released from rest and slides down a hillside that slopes downward at \(65.0^{\circ}\) from the horizontal. (a) What minimum value must the coefficient of static friction between the hill and ball surfaces have for no slipping to occur? (b) Would the coefficient of friction calculated in part (a) be sufficient to prevent a hollow ball (such as a soccer ball) from slipping? Justify your answer. (c) In part (a), why did we use the coefficient of static friction and not the coefficient of kinetic friction?

A large uniform horizontal turntable rotates freely about a vertical axle at its center. You measure the radius of the turntable to be \(3.00 \mathrm{~m} .\) To determine the moment of inertia \(I\) of the turntable about the axle, you start the turntable rotating with angular speed \(\omega\), which you measure. You then drop a small object of mass \(m\) onto the rim of the turntable. After the object has come to rest relative to the turntable, you measure the angular speed \(\omega_{\mathrm{f}}\) of the rotating turntable. You plot the quantity \(\left(\omega-\omega_{\mathrm{f}}\right) / \omega_{\mathrm{f}}\) (with both \(\omega\) and \(\omega_{\mathrm{f}}\) in rad \(\left./ \mathrm{s}\right)\) as a function of \(m\) (in kg). You find that your data lie close to a straight line that has slope \(0.250 \mathrm{~kg}^{-1}\). What is the moment of inertia \(I\) of the turntable?

A small \(10.0 \mathrm{~g}\) bug stands at one end of a thin uniform bar that is initially at rest on a smooth horizontal table. The other end of the bar pivots about a nail driven into the table and can rotate freely, without friction. The bar has mass \(50.0 \mathrm{~g}\) and is \(100 \mathrm{~cm}\) in length. The bug jumps off in the horizontal direction, perpendicular to the bar, with a speed of \(20.0 \mathrm{~cm} / \mathrm{s}\) relative to the table. (a) What is the angular speed of the bar just after the frisky insect leaps? (b) What is the total kinetic energy of the system just after the bug leaps? (c) Where does this energy come from?

A solid uniform sphere and a thin-walled, hollow sphere have the same mass \(M\) and radius \(R .\) If they roll without slipping up a ramp that is inclined at an angle \(\beta\) above the horizontal and if both have the same \(v_{\mathrm{cm}}\) before they start up the incline, calculate the maximum height above their starting point reached by each object. Which object reaches the greater height, or do both of them reach the same height?
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