Let it roll

A plank of mass M M is moving on a smooth horizontal surface with speed v v_{\circ} . At time t = 0 t=0 , a sphere of mass m m and radius r r is gently placed on it and simultaneously. A constant horizontal force F F is applied on the plank in the opposite direction of v v_{\circ} . Find the time at which sphere starts pure rolling on the plank . The coefficient of friction between the plank and sphere is μ \mu .


M = 4 k g M=4~kg , m = 2 k g m=2~kg , μ = 0.6 \mu=0.6 , g = 10 m / s 2 g=10~m/s^2 , v = 5 m / s v_{\circ}=5~ m/s , F = 40 N F=40~N .


The answer is 0.147.

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1 solution

Tanishq Varshney
Jan 4, 2016

F.B.D of sphere for no vertical acceleration N = m g N=mg

f = μ m g \Rightarrow f= \mu mg (as f = μ N f= \mu N )

a = μ g \Rightarrow a=\mu g

v = μ g t v=\mu gt , Also α = μ m g r I ω = μ m g r t I \alpha=\frac{\mu mgr}{I} \Rightarrow \omega =\frac{\mu mgrt}{I} where I I is moment of inertia.

The velocity of the lower most point of the sphere at time t t is

v s = v + ω r v_{s}=v+ \omega r

v s = μ g t + μ m g r 2 I t \large{v_{s}=\mu gt + \frac{\mu mg r^2}{I}t}

From F.B.D of plank

a p = ( F + f M ) = ( F + μ m g M ) \large{\Rightarrow a_{p}=- \left(\frac{F+f}{M} \right)=- \left(\frac{F+\mu mg}{M}\right)}

v p = v o ( F + μ m g M ) t \large{v_{p}=v_{o}- \left(\frac{F+ \mu mg}{M} \right)t}

For no slipping v s = v p v_{s}=v_{p}

μ g t + μ m g r 2 I t = v o ( F + μ m g M ) t \large{\Rightarrow \mu gt+ \frac{\mu mg r^2}{I}t=v_{o}- \left(\frac{F+ \mu mg}{M} \right)t}

as I = 2 5 m r 2 I=\frac{2}{5} mr^2

t = v o 7 2 μ g + ( F + μ m g M ) \Large{\Rightarrow t=\frac{v_{o}}{\frac{7}{2}\mu g+\left(\frac{F+\mu mg}{M} \right)}}

is not the question overrated.it is not a 400 points question.

aryan goyat - 5 years, 5 months ago

It didnt accpt 0.14 . What do you have to say to this ?

SOURAV MISHRA - 5 years, 4 months ago

Why won't the initial velocity of the plank have any effect on the velocity of the sphere, as soon as it is kept on it...?

A Former Brilliant Member - 5 years, 5 months ago

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W͙e͙l͙l͙ w͙h͙y͙ w͙o͙u͙l͙d͙ i͙t͙ h͙a͙v͙e͙.??????

SHASHANK GOEL - 5 years, 5 months ago

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I don't know. Cuz the the friction will restrict the motion of the lowest point and hence, move with some velocity.

A Former Brilliant Member - 5 years, 5 months ago

🅖🅞🅣 🅣🅗🅔 🅢🅐🅜🅔! 5/34

SHASHANK GOEL - 5 years, 5 months ago

Good :-)Another method would be angular momentum conservation!

A Former Brilliant Member - 5 years, 5 months ago

How will you explain the friction acts in backward direction for the sphere?

Md Zuhair - 3 years, 7 months ago

good question!!!! but very much overrated.

A Former Brilliant Member - 3 years, 6 months ago

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I accept the fact. Well it seems you are revising your 11th :)

Md Zuhair - 3 years, 6 months ago

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how do you know?

A Former Brilliant Member - 3 years, 6 months ago

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@A Former Brilliant Member You did this question..... thats why. And you are in XIIth . So i said :)

You can try my Mechanics Set for JEE. Its nice

Md Zuhair - 3 years, 6 months ago

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@Md Zuhair thanks!!! will try them soon.

A Former Brilliant Member - 3 years, 6 months ago

@A Former Brilliant Member Here you go

https://brilliant.org/profile/md-909vmr/sets/my-best-picks-for-jee-mechanics/

Md Zuhair - 3 years, 6 months ago

1 pending report

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