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Tanzania [10]
3 years ago
12

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 heigh

t 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.
Physics
1 answer:
PtichkaEL [24]3 years ago
8 0
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
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