Posted by **Dermot** on Saturday, August 7, 2010 at 2:46pm.

A uniform solid sphere of mass 4kg and diameter 20cm initally at rest, begins to roll without slipping under the influece of gravity, down an incline that makes an angle of 15degrees to the horizontal

1 calculate the angular acceleration of the sphere

2 what is the total kinetic energy of the ballafter 3s?

3 how far does it roll?

(assume I=(2/5)Mr^2 for a uniform solid sphere.)

- Physics -
**drwls**, Sunday, August 8, 2010 at 12:38am
1. Let a be the rate at which the sphere accelerates. After time T from the start of motion, the velocity is

V = a*T and the angular velocity is alpha*T

PE loss equals KE gain.

M g *(1/2)*a T^2*sin 15 = (7/10) M *(a*T)^2

(1/2)a*g*sin15 = (7/10)*a^2

a = (5/7)*g*sin15

The 7/10 comes from adding translational and rotational kinetic energy, using the I value for a solid sphere.

2. Compute V after 3 seconds using the acceleration from part 1. Then compute the total kinetic energy.

3. It rolls to the bottom. Don't you mean how far does it roll in 3 seconds?

- Physics -
**Dermot**, Sunday, August 8, 2010 at 10:50am
yes, sorry after 3 s, thanks.

- Physics -
**Damon**, Sunday, August 8, 2010 at 11:19am
constant acceleration a

d = (1/2) a t^2

by the way the angular acceleration alpha = a/R

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