Question

A small ball with mass$\mathbf{\text{1.30\hspace{0.17em}kg}}$is mounted on one end of a rod$\mathbf{\text{0.780\hspace{0.17em}m}}$ long and of negligible mass. The system rotates in a horizontal circle about the other end of the rod at $\mathbf{\text{5010\hspace{0.17em}rev/min}}$.

(a) Calculate the rotational inertia of the system about the axis of rotation.

(b) There is an air drag of${\mathbf{\text{2.30×10}}}^{\mathbf{\text{-2}}}\mathbf{\text{N}}$ on the ball, directed opposite its motion. What torque must be applied to the system to keep it rotating at constant speed?

Short answer

a) The rotational inertia of the system about the axis of rotation is $0.791\text{\hspace{0.17em}kg}.{\text{m}}^2$.

b) The torque that must be applied to the system to keep it rotating at constant speed is $1.79\times {10}^{-2}\text{Nm}$.

Step-by-step solution

Step 1: Given

i) Mass of the ball, $M=1.30\text{kg}$

ii) Length of rod, $r=0.780\text{\hspace{0.17em}m}$

iii) Drag force,$F=2.30\times {10}^{-2}\text{\hspace{0.17em}N}$

Determining the concept

Using the formula of rotational inertia about an axis of rotation, find it from the given mass and length of rod. Then, using the formula of torque, find the torque that must be applied to the system to keep it rotating at constant speed for the given air drag.

Formulae are as follow:

$I=M{r}^2$

$\tau =rF$

where, Iis moment of inertia, M, m are masses, r is radius,$\tau$is torque and F is force.

(a) Determining the rotational inertia of the system about the axis of rotation 

Forrotational inertia about an axis of rotation of rod,

$I=M{r}^2$

$I=(1.30\text{\hspace{0.17em}kg}){(0.780\text{m})}^2$

$I=0.791\text{\hspace{0.17em}kg}.{\text{m}}^2$

Hence, the rotational inertia of the system about the axis of rotation is $0.791\text{\hspace{0.17em}kg}.{\text{m}}^2$

(b) Determining thetorque that must be applied to the system to keep it rotating at constant speed 

For torque that must counteract the drag force,

$\tau =rF$

$\tau =(0.780\text{m})(2.30\times {10}^{-2}\text{\hspace{0.17em}N})$

$\tau =1.79\times {10}^{-2}\text{\hspace{0.17em}N.m}$

Hence,the torque that must be applied to the system to keep it rotating at constant speed is $1.79\times {10}^{-2}\text{\hspace{0.17em}N.m}$