Question

A block of mass m is suspended released from the top of a frictionless fixed hemisphere as shown.Find

i) the angle with the vertical where It Breaks off

ii) the velocity at the instant When It Breaks off

iii) the height where It Breaks off​

Answer 1

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At B ;N=0

Therefore,

mgcos$\theta$ = $\frac{M{v}^{2}_{B}}{R}$

Therefore,

${V}_{B}$ = $\sqrt {gRcos \theta}$______(1)

Now by equation of energy between A and B we have;

0+mgR= $\frac {1}{2}$ ${M{v}^{2}_{B}}$+ mgh

put ${V}_{B}$ from (1) and h=Rcos $\theta$

Therefore,

${V}_{B}$=$\large{\boxed{\boxed{\sqrt{\frac{2}{3}gR}}}}$ and h= ${\boxed{\boxed{\frac{2R}{3}}}}$ from the bottom

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Answer 2

Check the attachment!