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Please explain how to solve the image problem fully in clear and step wise ............. ..(

Sudhanva G V , 6 Years ago
Grade 11
anser 1 Answers
Samyak Jain
Let mass of particle be m, radius of semicircular track be R and acceleration due to gravity is g.
Weight of particle is mg.
At point X, let normal reaction of track be N. Resolve the weight (mg) vector along the radius (which is mgcos\theta) and tangent to the track (which is mgsin\theta). Since initial velocity is zero, net centripetal force N – mgcos\theta = 0.
The particle will descend the track due to mgsin\theta component.
We can apply energy conservation here because surface of semicircle is smooth, work done by normal reaction is zero and weight of the particle is conservative force i.e. there is no non-conservative force.
i.e. Loss in potential energy = Gain in kinetic energy            ….(1)
Clearly, vertical displacement of particle from X to Y is h = Rcos\theta – Rsin\phi = R(cos\theta – sin\phi).
\therefore loss in potential energy = mgh = mgR(cos\theta – sin\phi)               ….(2)
Let speed of the paricle at point Y be v. Also, its initial speed is zero.
Then gain in kinetic energy = (1/2)mv2 – (1/2)m02 =  (1/2)mv2     ….(3)
From (1),(2) & (3), mgR(cos\theta – sin\phi)  =  (1/2)mv2             ….(4)
Now, consider point Y. Resolve mg radially and tangentially.
Radial component of mg  = mg sin\phi  and  tangential component of mg = mg cos\phi.
It is given at point Y, normal reaction vanishes i.e. N = 0.
\therefore Net centripetal force is mg sin\phi = mv2 / R   \Rightarrow  mv2 = mgR sin\phi.
Substitute the value of mv2 in (4). We get
mgR(cos\theta – sin\phi)  =  (1/2)mgR sin\phi  \Rightarrow 2cos\theta – 2sin\phi = sin\phi
 2cos\theta  =  3sin\phi   or   sin\phi  =  (2/3) cos\theta
Last Activity: 6 Years ago
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