Physics

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.

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  3. 👁 301
  1. Its an angle of 30 not 300

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  2. 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

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    posted by Damon

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