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Physics / Physics with MasteringPhysics 4 / Chapter 11 / Problem 14P

Physics with MasteringPhysics | 4th Edition | ISBN: 9780321541635 | Authors: James S. Walker

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A wheel on a gameshow is given aninitiai angular speed of

Chapter 11, Problem 14P

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QUESTION:

Problem 14P

A wheel on a gameshow is given aninitiai angular speed of 1.22 rad/s. It comes to rest after rotating through 0.75 of a turn. (a) Find the average torque exerted on the wheel given that it is a disk of radius 0.71 m and mass 6.4 kg. (b) If the mass of the wheel is doubled and its radius is halved, will the angle through which it rotates before coming to rest increase, decrease, or stay the same? Explain. (Assume that the average torque exerted on the wheel is unchanged.)

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QUESTION:

Problem 14P

A wheel on a gameshow is given aninitiai angular speed of 1.22 rad/s. It comes to rest after rotating through 0.75 of a turn. (a) Find the average torque exerted on the wheel given that it is a disk of radius 0.71 m and mass 6.4 kg. (b) If the mass of the wheel is doubled and its radius is halved, will the angle through which it rotates before coming to rest increase, decrease, or stay the same? Explain. (Assume that the average torque exerted on the wheel is unchanged.)

ANSWER:

Step 1 of 2

The mathematical expression for torque is the product of moment of inertia and angular acceleration.

From the data given, we can calculate the moment of inertia and angular acceleration of the wheel and then the torque.

Part a

Initial angular speed $\omega{}{\ }_{i}=1.22$ rad/s

Final angular speed $\omega{}{\ }_{f}=0$ ( the wheel comes to rest )

The angular displacement $\theta{}=0.75$ turn

$\theta{}=0.75\times{}2\pi{}$ rad

$\theta{}=4.71$ rad

Let the angular acceleration be $\alpha{}$ .

From the equation,

${\omega{}{\ }_{f}}^{2}=\omega{}{{\ }_{i}}^{2}+2\alpha{}\theta{}$

$0=1.2{2}^{2}+2\alpha{}\times{}4.71$

$\alpha{}=\ -1.2{2}^{2}/(2\times{}4.71)$ rad/s2

$\alpha{}=\ -0.16$ rad/s2

The angular acceleration is -0.16 rad/s2. The negative sign signifies that the wheel finally comes to rest.

The mass of the wheel $m=6.4$ kg

The radius of the wheel $r=0.71$ m

The moment of inertia of the wheel is,

$I=\frac{1}{2}m{r}^{2}$

$I=\frac{1}{2}\times{}6.4\times{}(0.71{)}^{2}$ kg $\cdot{}$ m2

$I=1.61$ kg $\cdot{}$ m2

Therefore, the average torque is,

$\tau{}=\alpha{}I$

$\tau{}=\ -0.16\times{}1.61$ N $\cdot{}$ m

$\tau{}=\ -0.26$ N $\cdot{}$ m

Therefore, the average torque is -0.26 N $\cdot{}$ m.

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