ball rolling on a hemisphere.?

A hemisphere of radius R is fastened to a table. at the top of the hemisphere, a small ball of mass m is at rest. The ball is given a push tangential to the surface of the hemisphere, as shown in figure I, which sets it in motion.

a)what is the maximum initial speed that the ball can have without immediately leaving the surface of the hemisphere? do I go like this with the question (mVmax^2)/r=normal force- weight?

b)the initial speed of the ball is sqrt(2/3)Vmax. what is the initial angular momentum? will this formula work? l=RmVmax

c)let α be the angle of the radius with respect to the y-axis, as shown in figure II. write down the mechanical energy of the ball as a function of α. will this equation work Em=mgRcosα+0.5m(Vmax)^2

I really need help with the last two d)calculate the angle at which the ball leaves the surface. (hint: what is the contact force between the ball and the surface at this angle?)

e)with the expression for potential energy you found at part (c), calculate the net torque on the ball via τ=-dU/dα what is the maximum torque experienced by the ball before it loses contact?

2 Answers

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  • Anonymous
    3 months ago
    Best Answer

    a)

    (mVmax^2) / r = m g

    Vmax = sqrt(g / r)

    b)

    L = r m sqrt(2/3)Vmax

    L = m sqrt(2 g r / 3)

    c)

    Em = 1/2 m V^2 - m g r (1 - cosα)

    V^2 = [Vmax^2 + 2 g r (1 - cosα)]

    then

    Em = 1/2 m [Vmax^2 + 2 g r (1 - cosα)] - m g r (1 - cosα)

    Em = 1/2 m Vmax^2

    in effect kinetic energy increases but potential energy decreases by the same amount

    d)

    the ball leaves the surface when

    m V^2 / r = m g cosα

    m [Vmax^2 + 2 g r (1 - cosα)] / r = m g cosα

    m Vmax^2 / r + 2 m g (1 - cosα) = m g cosα

    m Vmax^2 / r + 2 m g - 2 m g cosα = m g cosα

    cosα = Vmax^2 / (3 g r) + 2/3

  • 3 months ago

    as shown in figure I ??

    • tazkir3 months agoReport

      Sorry forgot to add
      https://s.yimg.com/tr/i/f188cc05839b4ba7814d8aa1d009a21e_A.jpeg

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