The total energy at the top point is
mv²/2+mg2R.
It is equal to the energy at lower point.
Its height is
h=R-Rcosθ = R(1-cosθ) =>
the total energy is
mv₁²/2 +mgR(1-cosθ).
Use the law of conservation of energy.
mv²/2+mg2R = mv₁²/2 +mgR(1-cosθ).
v₁² = v²+2gR(1+cosθ)= ….
At the lower point
ma=N-mgcosθ
N=ma+ mgcosθ=
=mv₁²/R +mgcosθ=…..
A mass M of 7.00E-1 kg slides inside a hoop of radius R=1.30 m with negligible friction. When M is at the top, it has a speed of 5.47 m/s. Calculate the size of the force with which the M pushes on the hoop when M is at an angle of 39.0°.
1 answer