Question

A solid ball with a mass of 3.56 kg and a radius of 0.094 m starts from rest at the top of a 86.4 ⦠slope with a vertical height of 2.50 m. it proceeds to roll without slipping down the slope. what is the translational speed of the ball when it reaches the bottom? the acceleration of gravity is 9.81 m/s 2 . answer in units of m/s.

Answer 1

Let: m=1.50Kg, r=0.094m, h=2.50m, teta=86.4 and g=9.81m/s^2. The moment of inertia of a sphere is I=(2/5)*m*r^2, vi=0m/s, hf=0m, and the condition that a spherical object is rolling whitout slipping is w=V/r. So, by conservation of energy: mgh= (1/2)*((m*Vf^2)+(I*wf)). substituting, and clearing, Vf=((10/7)*g*h)^(1/2)=((10/7)*(9.81)*(2.50))^(1/2)=5.9190 m/s