Chapter 8.1, Problem 2PP

(a)

To determine

The angular displacement in given time.

Answer to Problem 2PP

The angular displacement in 3 minutes is $6\pi \text{radians}$

Explanation of Solution

Given:

Time taken by a rotating toy to complete one counterclockwise rotation in 1 min

Formula used:

One revolution is equivalent to $2\pi \text{radians}$

Calculation:

Angular displacement in 1 min is, $\theta =2\pi \text{or }360°$

So, the angular displacement in 3 min is,

  $\begin{array}{l}{\theta }_1=3\left(\omega \right)\hfill \\ =3\left(2\pi \right)\hfill \\ =6\pi \text{radians}\hfill \end{array}$

Conclusion:

Therefore, the angular displacement in 3 min is $6\pi \text{radians}$

(b)

To determine

The angular velocity of toy in rad/min.

Answer to Problem 2PP

The angular velocity of toy in rad/min is 6.28 rad/min

Explanation of Solution

Given:

Time taken by a rotating toy to complete one counterclockwise rotation in 1 min

Formula used:

Angular velocity is,

  $\omega =\frac{d\theta }{dt}$

Calculation:

Angular velocity,

  $\begin{array}{c}\omega =\frac{d\theta }{dt}\\ =\frac{2\pi }{1\min }\\ \text{=}\text{2π}\text{rad/min}\\ =\text{2}\times \text{3}\text{.14}\text{rad/min}\\ \omega \text{=}\text{6}\text{.28}\text{rad/min}\end{array}$

Conclusion:

Therefore, angular velocity of toy is6.28 rad/min

(c)

To determine

To identify: Whether the toy angular acceleration will be positive or negative if it is turned off.

Answer to Problem 2PP

The angular acceleration of toy is negative.

Explanation of Solution

Given:

Time taken by a rotating toy to complete one counterclockwise rotation in 1 min.

Formula:

Angular acceleration is,

  $\alpha =\frac{d\omega }{dt}$

Calculation:

If the toy is turned off, the angular velocity will reduce to 0. Final angular velocity = 0 rad/min

So, angular acceleration will be,

  $\begin{array}{l}\alpha =\frac{d\omega }{dt}\\ \text{=}\frac{\text{(Final angular velocity}-\text{Initial angular velocity)}}{\text{Time}}\\ \text{=}\frac{{\omega }_f-{\omega }_i}{t_f-t_i}\\ \text{Since}\text{after }\text{}\text{turning}\text{off}\text{of}\text{toy, }{\omega }_f\text{=}\text{0}\\ \alpha =\frac{-{\omega }_i}{t_f-t_i}\end{array}$

Now, $t_f-t_i$ cannot be negative, so the angular acceleration will be negative from the above expression.

Conclusion:

Therefore,if toy is turned off then there will be a decrease in its angular velocity. So angular acceleration of toy is negative.