Question

A hollow, spherical shell with mass 2.00kg rolls without slipping down a slope angled at 38.0?.

Answer 1

Answer:

$\mu = 0.31$

Explanation:

Given data:

mass = 2.00 kg

slope angle = 38.0

From figure

balancing force

$mgsin\theta - f = ma$   .....1

Balancing torque

$F_R = \frac{2}{3} mR^2 \alpha$ ......2

for pure rolling

$\alpha = \frac{a}{R}$

$F = \frac{2}{3} ma$

from 1 and 2nd equation

$mgsin\theta - \frac{2}{3}ma = ma$

$mgsin\theta = \frac{5}{3} ma$

$a = \frac{3}{5} g sin\theta$

 $= \frac{3\theta 9.8 sin 38}{5} = 3.62 m/s^2$

$F =\mu N$

   $= \frac{2}{3} ma$

   $= \frac{2}{3} 2\times 3.62 = 4.83 N$

N =normal force $= mgsin\theta = 2 \times 9.8 sin 38 = 15.44 N$

$\mu \times 15.44 = 4.83$

solving for  coefficent of friction we get

$\mu = 0.31$