Sriking the support

Classical Mechanics Level 5

A block of mass 0.1 kg is tied to one end of a light string of length 0.1m whose other end is fixed to a rigid support as shown in the figure. The block is given a speed v from the lower most position such that $\sqrt { 2 } <v<\sqrt { 5 }$ . When the block just leaves the circle the string is cut and the block moves as a projectile in a parabolic path and strikes the rigid support to which the string was attached. Find $\alpha$ if v= $\sqrt { \tan { \alpha } }$ where $\alpha$ is an acute angle in degrees.

This problem is originally part of set Mechanics problems by Abhishek Sharma .

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The answer is 75.

1 solution

Abhishek Sharma
Sep 19, 2014

answer answer

As at point A tension is equal to 0,

$mgcos\theta =\frac { m{ v }^{ 2 } }{ l }$

Using conservation of energy,

$\frac { 1 }{ 2 } m({ u }^{ 2 }-cos\theta )=mg(l(1+cos\theta ))$

Solving it we get,

$cos\theta =\frac { { u }^{ 2 }-2 }{ 3 }$

Now we form two equations using 2nd equation of motion and eliminate t to get,

$\tan ^{ 2 }{ \theta =2 }$

$\cos ^{ 2 }{ \theta =\frac { 1 }{ 3 } }$

Now replace $\cos { \theta }$ with the value which we obtained earlier to get,

$u=\sqrt { 2\quad +\sqrt { 3 } }$

$\alpha =\tan ^{ -1 }{ 2+\sqrt { 3 } }$

$\alpha =75°$

Can you please explain to me what is the 2nd equation of motion and how did you get cos^2 theta =1/3

Winston Cahya - 4 years, 11 months ago

You missed $v^{2}$ in the conservation of energy equation. :)

satvik pandey - 6 years, 4 months ago

It is very much there, $cos\theta$ term represents the $v^{2}$ term.

Abhishek Sharma - 6 years, 4 months ago

Sriking the support

The answer is 75.

1 solution

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