Question

You are spinning a rock, of mass 0.75 kg, at the end of a string of length 0.86m in a vertical circle in uniform circular motion. What minimum speed is necessary so that there is no tension in the string at the top of the circle but the rock stays in the same circular path? If the string's maximum tension is 45 N, what is the maximum speed the rock can have so that the string does not break? At what point in the vertical circle does this maximum value occur?

Answer 1

Answer:

v (minimum speed) = 2.90 m/sec.

$\\ \\ maximum speed (v)= 6.57 m/sec.$

Maximum value of speed will occur at lowest point of vertical circle.

Explanation:

a)  What minimum speed is necessary so that there is no tension in the string at the top of the circle but the rock stays in the same circular path?

Using the force balance expression at the top of the circle,

Gravitational Force + Tension force = Centrifugal force

$m*g + T = m*v^2/R$

Given that : T = 0

R = length of string = 0.86 m

mass of the spinning rock = 0.75 kg

$v = \sqrt{g*R}$

$v = \sqrt{9.81*0.86}$

v (minimum speed) = 2.90 m/sec.

b) what is the maximum speed the rock can have so that the string does not break?

Here the  force balance at bottom of circle is represented by the illustration:

$T = m*g + m*v^2/R$

Given that:

maximum tension T = 45 N

maximum speed v = ??

mass  m = 0.75 kg

∴

$45 - 0.75*9.81 = 0.75*\frac{v^2}{0.86} \\\\v^2 = 0.86*(45 - 0.75*9.81)/0.75 \\ v = \sqrt{0.86*(45 - 0.75*9.81)/0.75\\ maximum speed (v)= 6.57 m/sec.$

c)

At what point in the vertical circle does this maximum value occur?

Maximum value of speed will occur at lowest point of vertical circle.

This is so because  at the lowest point; the tension in string will be maximum.