A classical mechanics problem by A Former Brilliant Member

A uniform cylinder of radius R R is spun about its axis to an angular velocity ω 0 \omega_0 and then placed in a corner. The coefficient of friction between the wall (and floor) and the cylinder is μ \mu . How many radians does it spin through before stopping?

Assumptions and Details

  • μ = 3 \mu=\sqrt{3}
  • ω 0 = 2 π rad s 1 \omega_0 = 2 \sqrt{\pi}\si{\radian\per\second}
  • R = 0.25 m R=\SI{0.25}{\meter}
  • g = 10 m s 2 g=\SI{10}{\meter\per\second\squared}


The answer is 0.0665.

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1 solution

Swagat Panda
Aug 6, 2016

Given: \text{Given:}
μ = 3 ω 0 = 2 π R = 0.25 m g = 10 m s 2 \boxed{ \rightarrow \mu=\sqrt3 \\ \rightarrow\omega_0=2\sqrt{\pi} \\ \rightarrow R=\SI{0.25}{\meter} \\ \rightarrow g=\SI{10}{\meter\per\second\squared}}

Let \text{Let} m m be the mass of the cylinder, F 1 be the frictional force acting along the floor and F 2 be the force acting along the wall as \text{be the mass of the cylinder, } F_1 \text{ be the frictional force acting along the floor and } F_2 \text{ be the force acting along the wall as} shown in the following figure. \text{shown in the following figure.} Equating components of force and normal reaction horizontally and vertically and using the laws of friction we get: \text{Equating components of force and normal reaction horizontally and vertically and using the laws of friction we get:} N 1 + F 2 m g = 0 ( 1 ) F 1 N 2 = 0 ( 2 ) F 1 = μ N 1 ( 3 ) F 2 = μ N 2 ( 4 ) \\N_{ 1 }+F_{ 2 }-mg=0\quad \cdots (1) \\F_{ 1 }-N_{ 2 }=0\quad \cdots (2) \\F_{ 1 }=\mu N_{ 1 }\quad \cdots (3) \\F_{ 2 }=\mu N_{ 2 }\quad \cdots (4) Using (1), (2), (3) and (4) we get: \text{Using (1), (2), (3) and (4) we get:} F 1 = μ m g 1 + μ 2 and F 2 = μ 2 m g 1 + μ 2 \boxed{F_{ 1 }=\dfrac { \mu mg }{ 1+\mu^2 }} \text{ and } \boxed{F_2=\dfrac{\mu^2mg}{1+\mu^2}} Moment of inertia, I is equal to \text{Moment of inertia, I is equal to} I = m R 2 2 \boxed{I=\dfrac{mR^2}{2}} Net torque acting on the cylinder \text{ Net torque acting on the cylinder} τ n e t \tau_{net} is \text{is} ( α \alpha is the angular acceleration): \text{is the angular acceleration):} τ n e t = F 1 R F 2 R = I α I α = μ m g 1 + μ 1 + μ 2 α = 2 μ g ( 1 + μ ) R ( 1 + μ 2 ) \tau_{net}=-F_1R-F_2R = I\alpha \\ \Rightarrow I\alpha=\mu mg\dfrac{1+\mu}{1+\mu^2} \\ \Rightarrow \boxed{\alpha=\dfrac{-2\mu g\left(1+\mu \right) }{R\left(1+\mu^2\right)}} α and τ n e t are with a negative sign because the direction of ω 0 was assumed to be positive and the τ n e t and α acting are opposing it. \alpha \text{ and } \tau_{net} \text{ are with a negative sign because the direction of }\omega_0 \text{ was assumed to be positive and the } \tau_{net} \text{ and } \alpha \text{ acting are opposing it.}

Now, when the circular motion of the cylinder stops, \text{Now, when the circular motion of the cylinder stops,} ω t 2 = 0 = ω 0 2 + 2 α θ θ = ω 0 2 ( 1 + μ 2 ) R 4 μ g ( 1 + μ ) = 4 π × ( 1 + ( 3 ) 2 ) × 0.25 4 × 10 ( 3 + ( 3 ) 2 ) θ = 0.066 39 rad 0.066 5 rad \omega^2_t=0=\omega^2_0+2\alpha\theta \\ \Rightarrow \boxed{\theta=\dfrac{\omega^2_0\left(1+\mu^2\right)R}{4\mu g \left(1+\mu \right)}=\dfrac { 4\pi \times \left( 1+\left( \sqrt { 3 } \right) ^{ 2 } \right) \times 0.25 }{ 4\times 10\left( \sqrt { 3 } +\left( \sqrt { 3 } \right) ^{ 2 } \right)} } \\ \Rightarrow \boxed{\theta=\SI{0.06639}{\radian}\approx\SI{0.0665}{\radian}}

wow you really did it !!!!!! congrats !! first one to solve it!!

A Former Brilliant Member - 4 years, 10 months ago

Yeah! Same method !!

Aniket Sanghi - 4 years, 10 months ago

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me too how much time did u take to solve it ? i did it in exactly 1 min 53 sec

A Former Brilliant Member - 4 years, 10 months ago

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I dint see the clock , but ha may be 1 to 2 min .

Aniket Sanghi - 4 years, 10 months ago

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@Aniket Sanghi i just uploaded it try this : https://brilliant.org/problems/zero-latex-knowledge/?ref_id=1250028

A Former Brilliant Member - 4 years, 10 months ago

Thanks, it took a lot of time to solve and then finally post a solution. Nice problem

Swagat Panda - 4 years, 10 months ago

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is ashish barnawal ( from your centre) also on brillant ?

A Former Brilliant Member - 4 years, 10 months ago

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yes, he is, the topper of vsat this time and usually he is 2nd best after Piyush Kansal from Hisar. But he is new to it and doesn't use it as much as we do.

Swagat Panda - 4 years, 10 months ago

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@Swagat Panda okay , sorry i am talking about classes here but why don't you attend classes ?

A Former Brilliant Member - 4 years, 10 months ago

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@A Former Brilliant Member No, its okay, I told you I was ill and I have to watch video archives of 10 days, which I am doing from last few days.

Swagat Panda - 4 years, 10 months ago

Shubham, what is the percentage of people who have attempted the question, because I think this question deserves a level 4.

Swagat Panda - 4 years, 10 months ago

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