Question

A 41 g ball rolls around a 64-cm-diameter L-shaped track at 55 rpm. What is the magnitude of the net force that the track exerts on the ball? Rolling friction can be neglected.

Answer 1

Answer:

0.44 N

Explanation:

The net force (centripetal force) acting on the ball is:

$F=m\omega^2 r$

where

m = 41 g = 0.041 kg is the mass of the ball

$\omega =55 rev/min$ is the angular velocity

$r=\frac{L}{2}=\frac{64 cm}{2}=32 cm=0.32 m$ is the radius of the track

First we need to convert the angular velocity into rad/s:

$\omega=55 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=5.76 rad/s$

And now we can calculate the magnitude of the force:

$F=(0.041 kg)(5.76 rad/s)^2 (0.32 m)=0.44 N$

Answer 2

The magnitude of the centrifugal force that has a mass of 41 g, radius of 32 cm, and RPM of 55 is 0.44 newtons.

What is a centrifugal force?

The centrifugal force is a pseudo force in a circular motion that acts along the radius and is directed away from the center of the circular path.

The centrifugal force is given as

F = mrω²

Where m is mass, r is a radius, and ω is angular velocity.

A 41 g ball rolls around a 64-cm-diameter L-shaped track at 55 rpm.

m = 41 g = 0.041 kg

The radius will be

$\rm r = \dfrac{64}{2} = 32$

r = 32 cm = 0.32 m

The angular velocity will be

$\rm \omega = \dfrac{2 \pi N}{60}\\\\\omega = \dfrac{2 \pi 55}{60}\\\\\omega = 5.76$

ω = 5.76 radian per second

Then the centrifugal force will be

F = 0.041 × 0.32 × 5.76²

F = 0.041 × 0.32 × 33.1776

F = 0.435 ≅ 0.44 N

More about the centrifugal force link is given below.

https://brainly.com/question/4669833