Question

A solid 0.4550 kg ball rolls without slipping down a track toward a vertical loop of radius R = 0.6750 m . What minimum translational speed v min must the ball have when it is a height H = 1.021 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/s 2 for the acceleration due to gravity.

Answer 1

Answer:

u = 3.35 m/s

Explanation:

given,

mass , m = 0.455 kg  

R = 0.675 m

Height of Loop = 1.021 m

the speed required at the top of loop be v

equating the force vertically

$m g =\dfrac{mv^2}{r}$

$9.81 =\dfrac{v^2}{0.675}$

v² = 6.622

v = 2.57 m/s

Let the initial speed of ball be u

using conservation of energy

$\dfrac{1}{2}mu^2 + \dfrac{1}{2}I\omega^2 + m g h = \dfrac{1}{2}I\omega^2 + \dfrac{1}{2}mv^2+ m g (2 R)$

where, $I =\dfrac{2}{5}mr^2$

$\dfrac{1}{2}mu^2 + \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{u}{r})^2 + m g h = \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{v}{r})^2 + \dfrac{1}{2}mv^2+ m g (2 R)$

$0.7 u^2 + g H = 0.7 v^2 + g(2R)$

$0.7 u^2 +9.81 \times 1.021= 0.7\times 2.57^2 + 9.81 \times 2\times 0.675)$

0.7 u² = 7.85092

u² = 11.2156

u = 3.35 m/s

the initial  speed is 3.35 m/s