Asked by amena
A solid ball is released from rest and slides down a hillside that slopes downward at an angle 52.0 from the horizontal.What minimum value must the coefficient of static friction between the hill and ball surfaces have for no slipping to occur?
Answers
Answered by
drwls
Let u be the static friction foefficient.
Linear Acceleration = a
= M[g*sin52/M -M*g cos52*u]/M
= g*[sin52 - u cos52]
Angular acceleration = alpha = a/R
= g [sin52 -ucos52)/R
(if there is no slipping)
= Ff*R/I < u*M*g*sin52*R/[(2/5)M R^2]
= (5/2)*u*g*sin52/R
Ff is the static friction force, which has a maximum value when slipping occurs.
To avoid slipping for a solid sphere, at that 52 degree slope angle,
(5/2)*u*g*sin52/R > g [sin52 -ucos52)/R
g's and R's cancel.
u*[(5/2)sin52 + cos52] > sin 52
u > 1/[(5/2) + cot 52] = 0.305
Linear Acceleration = a
= M[g*sin52/M -M*g cos52*u]/M
= g*[sin52 - u cos52]
Angular acceleration = alpha = a/R
= g [sin52 -ucos52)/R
(if there is no slipping)
= Ff*R/I < u*M*g*sin52*R/[(2/5)M R^2]
= (5/2)*u*g*sin52/R
Ff is the static friction force, which has a maximum value when slipping occurs.
To avoid slipping for a solid sphere, at that 52 degree slope angle,
(5/2)*u*g*sin52/R > g [sin52 -ucos52)/R
g's and R's cancel.
u*[(5/2)sin52 + cos52] > sin 52
u > 1/[(5/2) + cot 52] = 0.305
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