A solid ball of inertia m rolls without slipping down a ramp that makes an angle θ with the horizontal.

Part A
What frictional force is exerted on the ball?
Express your answer in terms of some or all of the variables m, θ, and the acceleration due to gravity g.

Part B
As a function of θ, what coefficient of friction is required to prevent slipping?
Express your answer in terms of θ.

Respuesta :

Answer:

Part a)

[tex]f = \frac{2}{7}mgsin\theta[/tex]

Part b)

[tex]\mu = \frac{2}{7} gtan\theta[/tex]

Explanation:

Part a)

Force equation on the inclined plane is given as

[tex]mgsin\theta - f = ma[/tex]

now for torque equation of the ball

[tex]fR = \frac{2}{5}mR^2 ( \frac{a}{R})[/tex]

[tex]f = \frac{2}{5}ma[/tex]

now from above two equations

[tex]mg sin\theta - \frac{2}{5}ma = ma[/tex]

[tex]mg sin\theta = \frac{7}{5} ma[/tex]

[tex]a = \frac{5}{7} gsin\theta[/tex]

so frictional force is given as

[tex]f = \frac{2}{5}ma[/tex]

[tex]f = \frac{2}{5}m(\frac{5}{7}gsin\theta)[/tex]

[tex]f = \frac{2}{7}mgsin\theta[/tex]

Part b)

Also we know that in the normal direction of the motion we have

[tex]F_n = mgcos\theta[/tex]

so we have

[tex]f = \mu F_n[/tex]

[tex]\frac{2}{7} mg sin\theta = \mu (mg cos\theta)[/tex]

now we have

[tex]\mu = \frac{2}{7} gtan\theta[/tex]