Question

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

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

Step-by-step solution

 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

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:

$F_g=mg=(0.03kg)\times (9.81m/s^2)=0.2943N$

Centripetal force :

$F_c=\frac{m{v}^2}{r}=\frac{(0.03kg)\times {(1.256m/s)}^2}{0.2m}=0.2366N$

Now calculate resultant force F_net

_{$F_{net}=\sqrt{(F_c^2+F_g^2)}=\sqrt{{(0.2943N)}^2+{(0.2366N)}^2}=0.38N$}