Question

A solid 0.6950 kg ball rolls without slipping down a track toward a vertical loop of radius ????=0.8950 m . What minimum translational speed ????min must the ball have when it is a height H=1.377 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 ???? . Use ????=9.810 m/s2 for the acceleration due to gravity.

Answer 1

Answer:

The minimum transnational speed is 4.10 m/s.

Explanation:

Given that,

Mass of solid ball = 0.6950 kg

Radius = 0.8950 m

Height = 1.377 m

We need to calculate the minimum velocity of the ball at bottom of the loop to complete the track

Using formula velocity at lower point

$v_{min}=\sqrt{5gR}$

Put the value into the formula

$v_{min}=\sqrt{5\times9.8\times0.8950}$

$v_{min}=6.62\ m/s$

We need to calculate the velocity

Using conservation of energy

P.E at height +K.E at height = K.E at the bottom

$mgH+\dfrac{1}{2}mv^2=\dfrac{1}{2}m(\sqrt{5gR})^2$

$v^2=(\sqrt{5gR})^2-2gH$

$v^2=(6.62)^2-2\times9.8\times1.377$

$v^2=16.8352$

$v=\sqrt{16.8352}$

$v=4.10\ m/s$

Hence, The minimum transnational speed is 4.10 m/s.