Physics
18.08.2022 08:42
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A41 g ball rolls around a 64-cm-diameter l-shaped track at 55 rpm. what is the magnitude

A41 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.
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davidoj13
davidoj13
4,7(4 marks)

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

kaoris6368
kaoris6368
4,5(93 marks)

Part a)

K_t = 500 J

Part b)

K_r = 250 J

Part c)

K = 750 J

Explanation:

Part a)

As we know that cylinder is rolling without slipping

So we have

v = R\omega

now we have

v = 10 m/s

m = 10 kg

now we know that translational kinetic energy is given as

K_t = \frac{1}{2}mv^2

K_t = \frac{1}{2}(10)(10^2)

K_t = 500 J

Part b)

Now for rotational kinetic energy we know that

K_r = \frac{1}{2}I \omega^2

K_r = \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2

K_r = \frac{1}{4}mv^2

K_r = \frac{1}{4}(10)(10^2)

K_r = 250 J

Part c)

Total kinetic energy is given as

K = K_r + K_t

K = 500 + 250

K = 750 J

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