Chapter 9, Problem 1DQ

(a)

To determine

The relation $v=r\omega$ is valid or not if the angular acceleration of an object is not constant.

Explanation of Solution

The relation for displacement is,

$s=r\theta$                                              (I)

$s$ is displacement.

$r$ is the radius of circular path

$\theta$ is angular distance.

Relation $v=r\omega$ is derived from the equation (I).

The relation $s=r\theta$ doesn’t depend on whether angular acceleration is constant or not. Thus, if an object doesn’t have a constant acceleration it will not affect its velocity. Hence relation $v=r\omega$ is valid.

Conclusion:

The relation $v=r\omega$ is valid.

(b)

To determine

The relation $a_{\tan }=r\alpha$  is valid or not if the angular acceleration of an object is not constant.

Explanation of Solution

The expression for tangential acceleration in terms of angular acceleration is,

$a_{\tan }=r\alpha$

$a_{\tan }$ is tangential acceleration.

$\alpha$ is angular acceleration.

Tangential acceleration is possessed by the object when it moves along the curve. The angular acceleration also doesn’t affect it. Thus relation $a_{\tan }=r\alpha$ is valid.

Conclusion:

The relation $a_{\tan }=r\alpha$ is valid.

(c)

To determine

The relation $\omega ={\omega }_0+\alpha t$ is valid or not if the angular acceleration of an object is not constant.

Explanation of Solution

The expression for angular velocity is,

$\omega ={\omega }_0+\alpha t$.

${\omega }_0$ is initial angular velocity.

$t$ is the time.

$\omega$ is the final angular velocity.

The above expression is derived from the assumption that the angular acceleration is constant. Thus, relation $\omega ={\omega }_0+\alpha t$ is not valid.

Conclusion:

The relation $\omega ={\omega }_0+\alpha t$ is not valid.

(d)

To determine

The relation $a_{\tan }=r{\omega }^2$ is valid or not if the angular acceleration of an object is not constant.

Explanation of Solution

The expression for tangential acceleration in terms of angular velocity is,

$a_{\tan }=r{\omega }^2$

For an object that moves in a circular path then it has centripetal acceleration and it doesn’t depends on the whether angular acceleration is constant or not. Thus above relation is valid. Hence the relation $a_{\tan }=r{\omega }^2$ is valid.

Conclusion:

The relation $a_{\tan }=r{\omega }^2$ is valid.

(e)

To determine

The relation $K=\frac{1}{2}I{\omega }^2$ is valid or not if the angular acceleration of an object is not constant.

Explanation of Solution

The expression for kinetic energy is,

$K=\frac{1}{2}I{\omega }^2$                                   (II)

$K$ is kinetic energy.

$I$ is moment of inertia.

The equation (II) is derived from,

$K=\frac{1}{2}m{v}^2$

$m$ is mass.

Substitute $r\omega$ for $v$ in above expression to find $K$.

$\begin{array}{c}K=\frac{1}{2}m{\left(r\omega \right)}^2\\ =\frac{1}{2}m{r}^2{\omega }^2\\ =\frac{1}{2}I{\omega }^2\end{array}$

The relation $r\omega$ is valid for any acceleration. Thus $K=\frac{1}{2}I{\omega }^2$ is valid.

Conclusion:

The relation $K=\frac{1}{2}I{\omega }^2$ is valid.