Question

1. A small block of mass m slides in a vertical circle of radius R on the inside of a circular track. There is no friction between the track and the block. At the bottom of the block’s path, the normal force that the track exerts on the block has magnitude nb. (a) What is the velocity of the block at the bottom of its path? (b) Use conservation of energy to find the velocity of the block at the top of its path. (c) Find the normal force that the track exerts on the block at the top of its path.

Answer 1

Answer:

a) $v_{b} = \sqrt{\frac{R}{m}( n_{b} - mg)}$

b) $v_{t} = \sqrt{\frac{R}{m}( n_{b} - mg) - 4gR}$

c) $N_{t} = = n_{b} - 6mg$

Explanation:

NOTE: No values are provided in this exercise, any values of the given parameters can be substituted into the answers provided

Let the mass of the block = m

Radius  = R

Normal force acting on the block, N =  $n_{b}$

a) Calculate the velocity of the block at the bottom of its path, $v_{b}$

According to Newton's second law of motion,

$\sum F = ma\\\sum F = N - mg$

N - mg = ma

Since the block clides in a vertical circle, the acceleration, a, is a centripetal acceleration

$a = \frac{v_{b} ^{2} }{R}$

$n_{b} - mg = \frac{mv_{b} ^{2} }{R}$

$\frac{R}{m}( n_{b} - mg) = v_{b} ^{2} \\ v_{b} = \sqrt{\frac{R}{m}( n_{b} - mg)}$

b) Use conservation of energy to find the velocity of the block at the top of its path.

According to the law of conservation of energy

$K_{b} = K_{t} + PE_{t}$.................(1)

$PE_{t} = mgh$

The height of the block's path at the top, h = twice the radius

h = 2R

$PE_{t} = mg(2R)\\PE_{t} = 2mgR$

From equation (1)

$\frac{1}{2} mv_{b} ^{2} = \frac{1}{2} mv_{t} ^{2} + 2mgR\\\frac{1}{2} v_{b} ^{2} = \frac{1}{2} v_{t} ^{2} + 2gR\\\frac{1}{2} v_{b} ^{2} - 2gR= \frac{1}{2} v_{t} ^{2}$

Substituting $v_{b}$ into the equation above

$\frac{1}{2} (\sqrt{\frac{R}{m}( n_{b} - mg)})^{2} - 2gR = \frac{1}{2} v_{t} ^{2} \\\frac{R}{m}( n_{b} - mg) - 4gR = v_{t} ^{2}\\v_{t} = \sqrt{\frac{R}{m}( n_{b} - mg) - 4gR}$

c) Find the normal force that the track exerts on the block at the top of its path.

From the Newton's law of motion applied in part a

$N_{t}$ + mg = ma

$a = \frac{v_{t} ^{2} }{R}$

$N_{t} = \frac{mv_{t}^{2}} {R} - mg$

$N_{t} = \frac{m(\frac{R}{m}( n_{b} - mg) - 4gR) }{R} - mg\\N_{t} = (n_{b} - mg) - 4mg - mg\\N_{t} = = n_{b} - 6mg$