Question

A small object of uniform density rolls up a curved surface with an initial velocity v it reaches upto a maximum height of 3v²/4g with respect to the initial position. What is the object

Answer 1

From the law of conservation of energy,
mgh = 1/2mv² + 1/2 Iω²
I = mk²(where, k= radius of gyration) and, ω = v²/r²
by solving the eq.,
v²= 2gh/1+(k²/r²)
put 'h' = 3v²/4g in this eq. and by solving, you'll get,
k²/r² = 1/2
which means, given object is a DISC

Answer 2

Answer:

From the law of conservation of energy,

mgh = 1/2mv² + 1/2 Iω²

I = mk²(where, k= radius of gyration) and, ω = v²/r²

by solving the eq.,

v²= 2gh/1+(k²/r²)

put 'h' = 3v²/4g in this eq. and by solving, you'll get,

k²/r² = 1/2

which means, given object is a DISC