Chapter 8.2, Problem 33SSC

(a)

To determine

To Calculate: Torque applied to the wheel.

Answer to Problem 33SSC

  $\tau =0.423\text{ N}\text{.m}$

Explanation of Solution

Given:

Mass of solid wheel, $m=5.2\text{ kg}$

Radius of the wheel, $r=\frac{0.55}{2}=0.275\text{ m}$

Initial angular speed, ${\omega }_o=0$

Final angular speed, $\omega =12\text{ rev/s = 12}\times \text{2}\pi \text{ = 75}\text{.36 rad/s}$

Time taken, $t=35\text{ s}$

Formula used:

Torque:

  $\tau =I\alpha$

Where,

  $\tau =$ Torque.

  $I=$ Moment of inertia.

  $\alpha =$ Angular acceleration.

Calculation:

Angular acceleration can be calculated from the second equation of kinematics in rotational mechanics.

  $\begin{array}{c}\alpha =\frac{\omega -{\omega }_o}{t}\\ =\frac{75.36-0}{35}\\ =2.15{\text{ rad/s}}^2\end{array}$

Therefore, Torque:

  $\begin{array}{c}\tau =I\alpha \\ =\frac{1}{2}m{r}^2.\alpha \\ =\frac{1}{2}\times 5.2\times {\left(0.275\right)}^2\times 2.15\\ =0.4227\\ =0.423\text{ N}\text{.m}\end{array}$

Conclusion:

Torque applied to the wheel: $0.423\text{ N}\text{.m}$ .

(b)

To determine

To Calculate: Force exerted on the strap.

Answer to Problem 33SSC

  $F=1.54\text{ N}$

Explanation of Solution

Given:

Torque on the wheel, $\tau =0.423\text{ N}\text{.m}$

Radius of the wheel, $r=0.275\text{ m}$

Formula used:

Torque:

  $\tau =F.r$

Where,

  $F=$ Force.

  $r=$ Perpendicular distance to the force.

Calculation:

Plugging in the given values in the formula:

  $\begin{array}{c}0.423=F\times 0.275\\ F=1.54\text{ N}\end{array}$

Conclusion:

Force exerted on the strap: $1.54\text{ N}$ .