Question

A sphere of mass m and radius R is placed at rest on a plank of mass M which is placed on a smooth horizontal surface as shown in the figure. The coefficient of friction is μ. At t=0, horizontal velocity v is given to the plank. Find the time after which, the sphere starts pure rolling.

$(a) \: \frac{2Mv}{(7M \: + \: 2m) \mu{g}}$
$(b) \: \frac{Mv}{(7M \: + \: 2m) \mu{g}}$
$(c) \: \frac{2Mv}{(5M \: + \: 2m) \mu{g}}$
$(d) \: \frac{Mv}{(5 M \: + \: 2m) \mu{g}}$
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Answer 1

Answer:

velocity of center of mass is v & angular velocity of sphere is 0 initially... friction will act in backward direction ....now f = kmg (f = friction force) ma = kmg a=-kmg (a is retardation of center of mass)

now , torque = fR = I(alfa) (alfa = angular accleratio

alfa = fR/I

= 5kg/2R (Isphere = 2MR2/5)

now after time t let velocity of ceneter of mass is V then

V = U + at (initial linear velocity is v)

V = v - kgt ............1

let at this time angular velocity os W then

W = Wo + (alfa)t (initial angular velocity is 0)

W = 5kgt/2R ............2

now if pure rolling has started then V = WR

so , v - kgt = 5kgt/2

t = 2v/7kg