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A solid ball is released from rest and slides down a hillside that slopes downward at 65.0" from the horizontal

AWhat minimum value must the coefficient of static friction between the hill and ball surfaces have for no slipping to occur?
BWould the coefficient of friction calculated in part (a) be sufficient to prevent a hollow ball (such as a soccer ball) from slipping? Justify your answer.
CIn part (a), why did we use the coefficient of static friction and not the coefficient of kinetic friction?

Verified Answer :

Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline 
X: mgsin65 - F = mAx 
Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) 
Moment about center of mass: 
Fr = Iα 
Now Ax = rα 
and F = umgcos65 
mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) 
umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) 
ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) 
ugcos65 = 0.4(gsin65 - ugcos65) 
1.4ugcos65 = 0.4gsin65 
1.4ucos65 = 0.4sin65 
u = 0.4sin65/1.4cos65 
u = 0.613 

The minimum value of the coefficient of static friction between the hill and ball surfaces have for no slipping to occur is \(\rm mg\sin 65^o -F =m\ a_x\).

What is kinematics?

It is the study of the motion of a body without considering mass. And the reason behind them is neglected.

A solid ball is released from rest and slides down a hillside that slopes downward at 65° from the horizontal.

The component along the x-axis will be

\(\rm mg\sin 65^o -F =m\ a_x\)

The component along the y-axis will be

\(\rm N- mg\cos 65^o = 0\)

Moment about the center of mass will be

\(\rm F\ r = I \alpha\)

Now

\(\rm a_x = r\alpha\) and \(\rm F = umg \cos65^o\)

Then we have

\(\rm mg \sin 65^o - umg\cos 65^o = mr\alpha \\\\g\ain 65^o - ug\cos65^o=r\alpha\)

And

\(\rm umg \cos 65^o \ (r) = 0.4mr^2 \alpha \\\\ug\cos65^o \ (r) = 0.4 (r\alpha)\)

Then we have

\(\begin{aligned} ug\cos65^o (r) &= 0.4r (g\sin65^o - ugcos65^o)\\\\ug\cos65^o &= 0.4r(g\sin65^o - ugcos65^o)\\\\1.4 ug\cos65^o &= 0.4 g\sin65^o\\\\u &= \dfrac{0.4sin65^o}{1.4\cos65^o}\\\\u &= 0.613 \end{aligned}\)

More about the kinematics link is given below.

https://brainly.com/question/25638908

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