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Chapter 8: Q. 53 (page 201)

30 g ball rolls around a 40-cm-diameter L-shaped track, shown in FIGURE P8.53, at 60 rpm. What is the magnitude of the net force that the track exerts on the ball? Rolling friction can be neglected.
Hint: The track exerts more than one force on the ball

Short Answer

Expert verified

The the net force that the track exerts on the ball FNet=0.38 N

Step by step solution

01

 Step1: Given Information

Mass of ball = 30 g =0.030 kg
Diameter of track =40 cm =0.40m

So radius =0.2 m
Angular velocity of ball =60 rpm

02

Explanation

First calculate Centripetal force using mv2/r

where m= mass, v= velocity and r = radius.

We know that Gravitational force =mg

Gravitational force acting downward and and centripetal force towards center of track.

So, net force is their resultant of these two force.

Gravitational force:

Fg=mg=(0.03kg)×(9.81m/s2)=0.2943N

Centripetal force :

Fc=mv2r=(0.03kg)×(1.256m/s)20.2m=0.2366N

Now calculate resultant force Fnet

Fnet=(Fc2+Fg2)=(0.2943N)2+(0.2366N)2=0.38N

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