Question

A uniform sphere (I = 2/5 MR 2 ) rolls down an incline. (a) What must be the incline angle if the linear acceleration of the center of the sphere is to be 0.10 gb) For this angle, what would be the acceleration of a frictionless block sliding down the incline

Answer 1

Answer:

Part a)

$\theta = 8.05 degree$

Part b)

$a = 1.37 m/s^2$

Explanation:

As the uniform sphere is rolling down the inclined plane then the net force on the sphere is given as

$mg sin\theta - F_f = ma$

also we have torque equation on it

$F_f R = I\alpha$

for pure rolling

$a = R \alpha$

$F_f = \frac{Ia}{R^2}$

now we have

$mg sin\theta = ma + \frac{Ia}{R^2}$

now we have

$mg sin\theta = (m + \frac{2}{5}m)a$

$a = \frac{5}{7}g sin\theta$

now given that

$a = 0.10 g$

so we have

$0.10 g = \frac{5}{7} g sin\theta$

$sin\theta = 0.14$

$\theta = 8.05 degree$

Part b)

If the inclined plane is frictionless then the acceleration is given as

$a = g sin\theta$

$a = 9.8(0.14)$

$a = 1.37 m/s^2$

Answer 2

Answer:

Explanation:

Given

Moment of inertia of sphere is $I=\frac{2Mr^2}{5}$

Here Friction will Provide torque

thus $f_r\times r=I\times \alpha$

$f_r=\frac{I}{r^2}$

Also friction will oppose weight sin component

$mg\sin \theta -f_r=ma$

$a=\frac{g\sin \theta }{1+\frac{I}{mr^2}}$

$a=0.1 g=\frac{g\sin \theta }{1+\frac{I}{mr^2}}$

$\sin \theta =0.14$

$\theta =8.407^{\circ}$

(b)Considering $\mu$ the coefficient of friction

acceleration of block sliding down the incline

$a=g\sin \theta -\mu g\cos \theta$