Question

A solid 0.6350 kg ball rolls without slipping down a track toward a vertical loop of radius R=0.8950 m.

What minimum translational speed vmin must the ball have when it is a height H=1.329 m above the bottom of the loop in order to complete the loop without falling off the track?

Assume that the radius of the ball itself is much smaller than the loop radius R. Use g=9.810 m/s2 for the acceleration due to gravity.

Answer 1

Answer:

The minimum translation speed is 4.21 m/s.

Explanation:

Given that,

Mass of solid = 0.6350 kg

Radius = 0.8950 m

Height = 1.329 m

We need to calculate the speed of the ball

Using formula of centripetal force

$F=\dfrac{mv^2}{r}$

$mg=\dfrac{mv^2}{r}$

$v_{b}=\sqrt{rg}$

Put the value into the formula

$v_{b}=\sqrt{0.8950\times9.8}$

$v_{b}=2.961\ m/s^2$

We need to calculate the minimum translation speed

Using conservation of energy

$K.E_{i}+P.E_{i}=K.E_{f}+P.E_{f}$

$\dfrac{1}{2}mv_{i}^2+mg(H-2r)=\dfrac{1}{2}mv_{f}^2+0$

$v_{i}^2=\sqrt{v_{f}^2-(2g(H-2r))}$

Put the value into the formula

$v_{i}^2=(2.961)^2-(2\times9.8(1.329-2\times0.8950))$

$v_{i}^2=\sqrt{(2.961)^2-(2\times9.8(1.329-2\times0.8950))}$

$v_{i}=4.21\ m/s$

Hence, The minimum translation speed is 4.21 m/s.