Chapter 8, Problem 5STP

To determine

To Choose: The correct option.

Answer to Problem 5STP

B. $2.5{\text{rad/s}}^{\text{2}}$

Explanation of Solution

Given:

The mass of the hoop is $m=5.0\text{kg}$ .

The force exerted tangentially to the hoop, $F=25\text{N}$

The radius of the hoop is, $r=2.0\text{m}$

Formula used:

The torque can be obtained by:

  $\tau =Fr\sin \theta$

Where, F is the applied force, r is the distance of the point of application of force from the axis of rotation, and $\theta$ is the angle between F and r .

The angular acceleration can be written as

  $\alpha =\frac{{\tau }_{net}}{I}$

Where, ${\tau }_{net}$ is the net torque on a rotating object, and $I$ is the moment of inertia of the object.

Calculation:

The moment of inertia of hoop is $I=m{r}^2$ .

Then, the angular acceleration of the hoop can be written as,

  $\begin{array}{c}\alpha =\frac{{\tau }_{net}}{I}\\ =\frac{Fr\sin 90°}{m{r}^2}\\ =\frac{F}{mr}\end{array}$

Using given values in above,

  $\begin{array}{c}\alpha =\frac{F}{mr}\\ =\frac{25}{5\times 2}\\ =2.5{\text{rad/s}}^{\text{2}}\end{array}$

Conclusion:

Thus, the angular acceleration of the hoop is $2.5{\text{rad/s}}^{\text{2}}$ .