Question

The figure below gives the speed v versus time t for a 0.490 kg object of radius 6.50 cm that rolls smoothly down a 30

Answer 1

For a sphere of constant density having mass M and radius R, the moment of inertia is
I = (2/5) M R^2
The kinetic energy of a travelling and spinning sphere of uniform density is
K = (1/2) I w^2 + (1/2) M v^2
K = (1/2)(2/5) M R^2 w^2 + (1/2) M v^2
K = (1/5) M R^2 w^2 + (1/2) M v^2
Where w is the angular speed. And where v is the speed of the sphere's center of mass with respect to the surface on which the sphere is moving, which is equal to the tangential speed of the sphere's rotational "equator," which is in contact with the surface on which the sphere is moving, with respect to the sphere's center of mass.
w = v / R
K = (1/5) M R^2 w^2 + (1/2) M v^2
K = (1/5) M v^2 + (1/2) M v^2
K = 0.7 M v^2
K = 0.7 M R^2 w^2
For your sphere, M=0.490 kg and R=0.065 m, so...
K = (0.00145 kg m^2) w^2
On a slope 30 degrees from horizontal, the acceleration from gravity would be
g = (9.8 m/s^2) cos(30) = 4.9 m/s^2
The work done by a ball rolling one meter along the slope is
E = M g (1 meter) = 2.401 Joules
K = E
The angular speed of the sphere after rolling 1 meter is
w = { K / (0.7 M R^2) }^(1/2)
w = { 2.401 Joules / (0.00145 kg m^2) }^(1/2)
w = 40.69 s^-1
v = wR = 2.644m/s
The average acceleration of the ball is found from the regular kinematic formula.
Vf^2 - V0^2 = 2 a dx
Where dx is the one meter measured along the slope, V0=0, Vf=v. The average acceleration, therefore, is
a = v^2/(2 meters) = 3.49 m/s^2
So the speed of a sphere of constant density having mass 0.490 kg and radius 0.065 meters, rolling freely down a 30 degree frictionless slope, is
v = (3.49 m/s^2) t