Question

A hollow sphere of mass m = 0.35 kg and radius r = 64 cm rolls along a flat surface at an initial speed of v, and then up a curved ramp with radius of curvature 4.7 m without slipping, until it reaches a maximum angle of 12 degrees around the curve and starts to roll backward. What was the initial speed of the ball in units of meters/second?

Answer 1

So, Initial Kinetic energy of hollow sphere is

K = -1 mv2 Iw 2

So,

$K=\frac{1}{2}mv^2+\frac{1}{2}\frac{7}{5}mr^2\frac{v^2}{r^2}$

so,

$K=\frac{6}{5}mv^2$

and when it reaches it's maximum angle then all the kinetic energy is converted in Potential energy

so,

U = mgR(1 - cose

R is radius of curvature,

theta = 12 deg

so,

from energy conservation

$\frac{6}{5}mv^2=mgR(1-cos\theta)$

so,

$v=\sqrt\frac{5mgR(1-cos\theta)}{6}$

so plug in all values we get initial velocity