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91Ó°ÊÓ

A small ball with mass1.30 k²µis mounted on one end of a rod0.780″¾ long and of negligible mass. The system rotates in a horizontal circle about the other end of the rod at 5010 r±ð±¹/³¾¾±²Ô.

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

(b) There is an air drag of2.30×10-2N on the ball, directed opposite its motion. What torque must be applied to the system to keep it rotating at constant speed?

Short Answer

Expert verified

a) The rotational inertia of the system about the axis of rotation is 0.791 k²µ.m2.

b) The torque that must be applied to the system to keep it rotating at constant speed is 1.79×10−2 Nm.

Step by step solution

01

Step 1: Given

i) Mass of the ball, M=1.30 kg

ii) Length of rod, r=0.780″¾

iii) Drag force,F=2.30×10−2 N

02

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=Mr2

Ï„=rF

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

03

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

Forrotational inertia about an axis of rotation of rod,

I=Mr2

I=(1.30 k²µ)(0.780 m)2

I=0.791 k²µ.m2

Hence, the rotational inertia of the system about the axis of rotation is 0.791 k²µ.m2

04

(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,

Ï„=rF

τ=(0.780 m)(2.30×10−2 N)

Ï„=1.79×10−2 N.³¾

Hence,the torque that must be applied to the system to keep it rotating at constant speed is 1.79×10−2 N.³¾

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