Answer to Question #145241 in Mechanics | Relativity for Aobakwe

Question

A 1916 kg solid sphere with a radius of 1.2 m rolls without slipping down an incline of θ = 29o. If the sphere makes 8 full revolutions in 5.5 seconds.

The rotational inertia of the sphere is?

The frictional force between the incline and the sphere is?

The speed of the center of mass of the Sphere is?

The distance S after 8 full revolutions is?

Answer 1

M = 1916 kg

R = 1.2 m

a. The rotational inertia

$I = \frac{2}{5}mR^2$

$I = \frac{2}{5} \times 1916 \times 1.2^2$

I = 1103.6 kgm²

b. $fR = I\frac{gsinθ}{R}$

$f = \frac{I \times g \times sin(29º)}{R^2}$

$f = \frac{1103.6 \times 9.8 \times 0.4848}{1.2^2} = 3641.14 \;N$

c. Speed

$v = w \times R$

The angular velocity

$w = \frac{2π}{T}$

$T = \frac{5.5}{8} = 0.687 \;s$

$w = \frac{2π}{0.687} = 9.14 \;rad/s$

$v = 9.14 \times 1.2 = 7.61 \;m/s$

d. Distance

$S = v \times 5.5 \;s = 7.61 \times 5.5 = 1.38 \; m$