Question

A uniform solid sphere with a mass of M = 390 grams and a radius R = 19.0 cm is rolling without slipping on a horizontal surface at a constant speed of 4.00 m/s. It then encounters a ramp inclined at an angle of 15.0 degrees with the horizontal, and proceeds to roll without slipping up the ramp. Use g = 10.0 m/s2. How far does the sphere travel up the ramp (measure the distance traveled along the incline) before it stops for an instant?

Answer 1

Answer:

l = 4.33 m

Explanation:

given,

mass of solid sphere = 390 gram = 0.39 kg

radius = R = 19 cm = 0.19 m

rolling with constant speed = 4 m/s

angle with horizontal = 15°

acceleration due to gravity = 10 m/s²

using energy conservation

$\dfrac{1}{2}I\omega^2 + \dfrac{1}{2}mv^2 = mgh$

I for sphere

$I = \dfrac{2}{5}mr^2$         v = r ω

$\dfrac{1}{2}\ \dfrac{2}{5}mr^2\times \dfrac{v^2}{r^2} + \dfrac{1}{2}mv^2 = mgh$

$\dfrac{7}{10}mv^2 = mgh$

$h = \dfrac{0.7 v^2}{g}$

$h = \dfrac{0.7\times 4^2}{10}$

h = 1.12 m

$l=\dfrac{h}{sin 15^0}$

$l=\dfrac{1.12}{sin 15^0}$

l = 4.33 m

the sphere will travel 4.33 m on the ramp.