Question

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

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.