A classical mechanics problem by Kenan Saracevic

Classical Mechanics Level 5

A body of mass m moves with constantly velocity v on a circle in vertical plane. What is the absolute work done by the friction force (which has constant k) during one rotation.

Take: m = 1kg v = 1 m/s k = 0,1

HINT: Try to do it without using calculus. :)


The answer is 0.63.

Kenan Saracevic by Kenan Saracevic , 23, Bosnia-Herzego…

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1 solution

Kelvin Hong
Aug 19, 2017

I am going to use calculus to solve this problem ^_^

As picture above states , T T stands for the force do by the circle route.

So

T − m g c o s θ = m v 2 r T-mgcos\theta = \frac{mv^2}{r}

T = m g c o s θ + m v 2 r T=mgcos\theta +\frac{mv^2}{r}

And the friction f f can be expressed by

f = k T = m g k c o s θ + m k v 2 r f=kT=mgkcos\theta + \frac{mkv^2}{r}

Using Δ W = f Δ s = f r Δ θ \Delta W = f\Delta s =fr\Delta {\theta}

Work can be calculated by using

W = ∫ 0 2 π f r d θ W= \int_0^{2\pi} frd\theta

W = ∫ 0 2 π m g r k c o s θ + m k r 2 d θ W= \int_0^{2\pi} mgrkcos\theta + mkr^2 d\theta

Everything except θ \theta are constants. By simple integration,

W = 2 π m k v 2 = 0.2 π = 0.628 W=2\pi mkv^2 =0.2 \pi =\boxed{0.628}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Can you please explain what T T is? And why is the frictional force equal to k T kT ? And since the question is highly unclear about what it demands, what surface is exerting the frictional force, is it a circular surface as in your diagram?

SUDHIR KUMAR - 2 years, 3 months ago

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