Block sliding off a table

Classical Mechanics Level 2

A small block slides on a horizontal friction-less surface $AB$ of a table with speed $\frac { 1 }{ 2 } \sqrt { gR }$ . At some point $B$ , the track becomes circular with radius $R$ (look at the diagram). The block leaves the table at a certain point $C$ . Determine the angle $\alpha$ .

\cos ^{ -1 }{ \frac { 5 }{ 6 } }

\sin ^{ -1 }{ \frac { 3 }{ 4 } }

\tan ^{ -1 }{ \frac { 3 }{ 4 } }

\cos ^{ -1 }{ \frac { 3 }{ 4 } }

2 solutions

Ajit Athle
May 22, 2014

Velocity at point B = (1/2)√(gR) Let’s say the block leaves the track at height h below B. h = R – Rcos(α) and mgcos(α) = mV²/R ----------(1), V being the velocity at the point of leaving the track when the radial component of weight fails to restrain the centrifugal force. Energy consideration gives us: (1/2)mV² =mg[R – Rcos(α)] + (1/2)mgR/4 Or V²/R = 2g(1 – cos(α)) + g/4. Plugging this value in (1), we get: 2g(1 –cos(α)) + g/4 = gcos(α) which, in turn, yields 3gcos(α) = 2g + g/4 = 9g/4 or cos(α) = 3/4

Jaivir Singh
Jun 9, 2014

COS OF ANGLE IS EQUAL TO VSQUARE PLUS 2RG UPON 3RG

Block sliding off a table

2 solutions

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