A uniform solid sphere rolls along a horizontal frictionless surface at 35 m/s and makes a smooth transition onto a frictionless incline having an angle of 300. How far up the incline does the sphere roll when it comes to a momentary stop? Note for a sphere I= 2/5Mr2, where M is the mass of the sphere and r is the radius of the sphere.
2 answers
Its an angle of 30 not 300
Original ke = (1/2)I omega^2 + (1/2)m v^2
omega = v/r = 35/r
ke=(1/2)(2/5)mr^2(35^2)/r^2+(1/2)m (35)^2
= (1/2)m [(2/5)+1](35)^2
= (1/2) m (1.4)(35)^2
distance up = h = L sin 20 = L/2
potential energy stopped at top = original ke
m g h = m g L/2 = (1/2)m(1.4)(35)^2
note m cancels
g = 9.81 on earth
omega = v/r = 35/r
ke=(1/2)(2/5)mr^2(35^2)/r^2+(1/2)m (35)^2
= (1/2)m [(2/5)+1](35)^2
= (1/2) m (1.4)(35)^2
distance up = h = L sin 20 = L/2
potential energy stopped at top = original ke
m g h = m g L/2 = (1/2)m(1.4)(35)^2
note m cancels
g = 9.81 on earth
2 answers