Respuesta :
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
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 [tex]\rm mg\sin 65^o -F =m\ a_x[/tex].
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
[tex]\rm mg\sin 65^o -F =m\ a_x[/tex]
The component along the y-axis will be
[tex]\rm N- mg\cos 65^o = 0[/tex]
Moment about the center of mass will be
[tex]\rm F\ r = I \alpha[/tex]
Now
[tex]\rm a_x = r\alpha[/tex] and [tex]\rm F = umg \cos65^o[/tex]
Then we have
[tex]\rm mg \sin 65^o - umg\cos 65^o = mr\alpha \\\\g\ain 65^o - ug\cos65^o=r\alpha[/tex]
And
[tex]\rm umg \cos 65^o \ (r) = 0.4mr^2 \alpha \\\\ug\cos65^o \ (r) = 0.4 (r\alpha)[/tex]
Then we have
[tex]\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}[/tex]
More about the kinematics link is given below.
https://brainly.com/question/25638908
