What Range of Speeds Can an Object Have Before a String Breaks?

[shawli]

Homework Statement

A light string can support a stationary hanging load of 25.0kg before breaking. An object of mass m = 3.00kg attached to the string rotates on a frictionless, horizontal table in a circle of radius r = 0.800m, and the other end of the string is held fixed. What range of speeds can the object have before the string breaks?

Homework Equations

F = (m*v^2)/r

The Attempt at a Solution

I seem to only be able to solve this question in terms of F. As in, the max velocity can be:

v = (F*0.800/25)^0.5

Yet there is an actual numeric answer to this question.

I'm not sure what to do with the '3.0kg' given. Any hints would be appreciated!

 

[dacruick]

when you spin this 3kg mass, there is a certain amount of tension in the string. The tension in the string cannot exceed 25kg * gravity. So what is the maximum speed you can spin a mass of 3kg so the tension in the string doesn't exceed 25kg * gravity?

 

[shawli]

Ohh, this is the definition of "hanging load"? I understand now! Thank you.

 

[dacruick]

Ohh, this is the definition of "hanging load"? I understand now! Thank you.

You're very welcome

 

[asteriatic]

I would first clarify any uncertainties or ambiguities in the given information. For example, it is not specified whether the hanging load of 25.0kg is the maximum weight the string can support or if it is the weight of the object attached to the string. Additionally, it is not specified if the object's mass of 3.00kg is included in the 25.0kg maximum weight. These details could affect the calculation and should be clarified.

Assuming the 25.0kg is the maximum weight the string can support and does not include the object's mass, the equation F = (m*v^2)/r can be used to solve for the maximum velocity before the string breaks. Plugging in the given values, we get:

25.0kg = (3.00kg*v^2)/0.800m

Solving for v, we get v = 6.32 m/s. Therefore, the range of speeds the object can have before the string breaks is 0 m/s < v < 6.32 m/s.

If the 25.0kg includes the object's mass, the equation would be slightly different:

25.0kg = (m*v^2)/0.800m

Solving for v, we get v = 7.75 m/s. Therefore, the range of speeds the object can have before the string breaks is 0 m/s < v < 7.75 m/s.

It is important to note that these are maximum velocities and the actual velocity of the object should be kept well below these values to ensure the string does not break. Additionally, the string may have a maximum tension limit as well, which should also be considered in this scenario.