Question

A solid ball is released from rest and slides down a hillside that slopes downward at an angle 55.0?from the horizontal.

What minimum value must the coefficient of static friction between the hill and ball surfaces have for no slipping to occur?

kmin=?

Answer 1

The moment of inertia of a uniform solid cylinder is, 1-2 MR The expression for the co efficient of the static friction is /tan α 2 tan a 2 -tan (55) 0.408 The coefficient of static friction between the hill and ball surfaces for no slipping to occur is 0.408

Answer 2

Erce Body dingzawn af Bll

By free body diagram of ball,

apply newton's law in y direction,

Fy = m*ay

given = ay = 0

Fy = N - m*g*cos$\theta$ = 0

N = m*g*cos$\theta$

apply newton's law in x direction,

Fx = m*ax

Fx = m*g*sin$\theta$ - fr = m*ax

here, fr = $\mu$s*N = $\mu$s*m*g*cos$\theta$

ax = g*(sin$\theta$ - $\mu$s*cos$\theta$) -eq(1)

Now by net torque balance on ball,

$\tau$net = I*$\alpha$

here, I = moment of inertia of ball = (2/5)*m*R^2

For no slipping $\alpha$ = ax*R

also, $\tau$net = fr*R

So, fr*R = [(2/5)*m*R^2]*(ax/R)

$\mu$s*m*g*cos$\theta$ = (2/5)*m*ax

from eq(1)

$\mu$s = (0.4*g*(sin$\theta$ - $\mu$s*cos$\theta$))/(g*cos$\theta$)

$\mu$s = (0.4*(sin$\theta$ - $\mu$s*cos$\theta$))/cos$\theta$

given, $\theta$ = 60.0 deg

So, $\mu$s = 0.4*tan(60.0 deg) - 0.4*$\mu$s

$\mu$s = 0.4*tan(60.0 deg)/(0.4 + 1)

$\mu$s = 0.495

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