A classical mechanics problem by Kushal Patankar

Classical Mechanics Level 4

A small ball of mass $m$ is attached to the end of the string of length $l=1 \ m$ whose other end is fixed. From its lowest position, the ball is given kinetic energy $\frac{mgl}{5}$ . Calculate the net acceleration $(in m/s^2)$ of the ball at the instant when the string makes an angle $\theta=37°$ with the vertical.

The answer is 6.

3 solutions

Abhishek Sharma
Apr 12, 2015

Using energy conservation,

Loss in KE = gain in GPE

${K}_{i}-\frac{1}{2}m{v}^{2}=mgl(1-\cos{\theta})$

$v=0$

${a}_{r}=\frac{{v}^{2}}{l}=0$

${a}_{t}=g\sin{\theta}$

$\boxed{a=6}$

Sir, Shouldn't it be mentioned in the question that, we will assume sin(37) = (3/5) & the "g" = 10?

Rubayet Tusher - 6 years, 1 month ago

Yes, it should be. However, even if you take exact values of $g$ and $\sin{37}$ the answer comes out to be $5.89$ which is close enough to $6$ .

Abhishek Sharma - 6 years, 1 month ago

@Abhishek Sharma THE BALL DOES NOT EVEN GO TO ANGLE 37 DEGREES IF g IS 9.8

William G. - 4 years ago

Net acc will be 10...

DragonBall Showdown - 3 years, 3 months ago

Its asking for net acc not tangential...

DragonBall Showdown - 3 years, 3 months ago

Rakshith Lokesh
Mar 19, 2018

by energy conservation mgl(1-cos(theta))+mgl/5= 1/2mv^2
make v square as subject .. v^2=2gl(6/5-cos(theta)) diffrentiate both sides 2v.dv/dx=2glsin(theta) {v.dv/dx=a} 2.a=2glsin(theta) put theta =45 degree and solve for a to get answer as 6.

Sushovan Haldar
Apr 12, 2015

velocity is 0.hence acclrtn is 6

A classical mechanics problem by Kushal Patankar

The answer is 6.

3 solutions

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