A solid metal ball starts from rest and rolls without slipping a distance of d =

Question

A solid metal ball starts from rest and rolls without slipping a distance of d = 4.6 m down a = 25° ramp. The ball has uniform density, a mass M = 4.5 kg and a radius R = 0.21 m.

1-Of the total kinetic energy of the sphere, what fraction is translational?

2-What is the translational kinetic energy of the sphere when it reaches the bottom of the incline?

3-What is the translational speed of the sphere as it reaches the bottom of the ramp?

4-What is the magnitude of the frictional force on the sphere?

5-Now let's change the problem a little. Suppose now that there is no frictional force between the sphere and the incline. Now, what is the translational kinetic energy of the sphere at the bottom of the incline?

Explanation / Answer

1. translational KE = mv^2 /2

rotational KE = I w^2 /2 = (2 m r^2 / 5) ( v /r)^2 /2 = mv^2 / 5

total KE = mv^2 /2 + mv^2 //5   = 7mv^2 / 10

so ,   trans KE / total KE = ( mv^2 /2) / (7mv^2 /10)   = 10/14   = 0.714

2. using energy conservationm,

mgh = 7mv^2 /10

9.81 x 4.6 x sin25 = 7v^2 / 10

v = 5.22 m/s

trans KE = mv ^2 /2 = 4.5 x 5.22^2 / 2 = 61.31 J

3. v = 5.22 m/s

4. using v^2 - u^2 = 2ad

5.22^2 - 0 =2 x a x 4.6

a = 2.96 m/s^2

along the inclne,

mgsin25 -f = ma

4.5x9.81 x sin25 - f    = 4.5 x 2.96

f = 5.34 N

5. if there is no friction,.

then sphere will gainonly translational KE .

at bottom rotational KE will be zero.

so total KE = mv^2 / 2

using energy conservation ,

KE = mgh = 4.5 x 9.81 x 4.6 x sin25 = 85.82 J

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