Rotational momentum conservation

[eeriana]

b]1. Homework Statement [/b]
A small block of mass .250kg is attached to a string passing through a hole in a frictionless horizontal surface. The block is originally revolving in a circle with a radius of .800 m about the hole with a tangential speed of 4 m/s. The string is then pulled slowly from below, shortening the radius of the circle in which the block revolves. The breaking strength of the string is 30N. What is the radius of the circle when the string breaks?

Homework Equations

KE = 1/2mv^2
MVoRo=MVR

The Attempt at a Solution

I am not sure how to get started. I know that momentum must be conserved.

This is what I did from a similar problem in a book. I am not sure if it is right.
1/2mv^2(Ro^2/R^2) and I got r=.57m If it is wrong can someone point me in the right direction.

Thanks.

 

[rl.bhat]

The breaking strength of the string is 30N. Breaking strength is due to centrfugal reaction and is equal to mv^2/R. From this find v^2/R.
According to conservation of angular momentum we have MVoRo=MVR
or VoRo=VR. Squaring both the sides we get Vo^2Ro^2 = V^2R^2. To the right hand side multiply and divide by R. So we have Vo^2Ro^2 = V^2R^3/R = (V^2/R.)R^3.Substitue the appropreate values and find R

 

[ogb p]

I can provide a response to this question by explaining the concept of rotational momentum conservation. Rotational momentum conservation is a fundamental principle in physics that states that the total angular momentum of a system is conserved in the absence of external torques. This means that the total angular momentum before an event or interaction is equal to the total angular momentum after the event or interaction.

In this specific problem, we can apply the principle of rotational momentum conservation to determine the radius of the circle when the string breaks. Initially, the block is revolving in a circle with a radius of 0.8m and a tangential speed of 4 m/s. This means that the initial angular momentum of the block is given by:

L = mvr = (0.250kg)(4m/s)(0.8m) = 0.8 kgm^2/s

When the string is pulled and the radius of the circle is shortened, the tangential speed of the block increases to maintain the same angular momentum. This can be seen from the equation for angular momentum, L = mvr, where v and r are inversely proportional. As the radius decreases, the tangential speed must increase to maintain the same value of angular momentum.

The maximum tangential speed that the block can reach before the string breaks is limited by the breaking strength of the string, which is given as 30N. This means that the maximum tension in the string can be 30N, which is equal to the centripetal force acting on the block. Using the equation for centripetal force, F = mv^2/r, we can solve for the maximum tangential speed:

30N = (0.250kg)v^2/r

v = √(30N * r / 0.250kg)

Now, we can substitute this value of v into the equation for angular momentum:

0.8 kgm^2/s = (0.250kg)(√(30N * r / 0.250kg))(r)

0.8 kgm^2/s = √(7.5Nr)

Squaring both sides and solving for r, we get:

r = 0.57m

Therefore, the radius of the circle when the string breaks is 0.57m. This can also be confirmed by plugging this value into the equation for centripetal force, F = mv^2/r, and solving for