A solid sphere of uniform density starts from rest and rolls without slipping a

Question

A solid sphere of uniform density starts from rest and rolls without slipping a distance of d = 3.2 m down a = 34°incline. The sphere has a mass M = 3.4 kg and a radius R = 0.28 m.

1)

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

KE tran/KEtotal =

2)

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

KE tran =

J

3)

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

v =

m/s

4)

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?

KE tran =

J

A 15 kg uniform disk of radius R = 0.25 m has a string wrapped around it, and a m = 3.2 kg weight is hanging on the string. The system of the weight and disk is released from rest.

1. If the system started from rest, how far has the weight fallen?

h fall =

Explanation / Answer

Here ,

1)

for rolling without slipping

v = r * w

Using conseravtion of energy

m * g * d * sin(theta) = 0.5 * I * w^2 + 0.5 m v^2

m * g * d * sin(theta) = 0.5 * 0.4 m * r^2 * (v/r)^2 + 0.5 m v^2

9.8 * 3.2 * sin(34) = 0.5 * (1.4 * v^2)

solving for v

v = 5.01 m/s

KEtrans/KEtotal = 0.5 m v^2/(m * g * d * sin(theta))

KEtrans/KEtotal = 0.5 * 5.01^2/(9.8 * 3.2 * sin(34))

KEtrans/KEtotal = 0.716

b)

KEtrans = 0.5 * 3.4 * 5.01^2

KEtrans = 42.7 J

c)

v = 5.01 m/s

4)

if there is no friction , all the energy will be translational kinetic energy

KEtrans = m * g * d * sin(theta)

KEtrans = 9.8 * 3.2 * sin(34) * 3.4

KEtrans = 59.62 J

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