String tension with an attached whirling mass

[MZA]

Homework Statement

A 0.40 kg mass, attached to the end of a .75 m string, is whirled around in a circular horizontal path. If the maximum tension that the string can withstand is 450 N, then what maximum speed can the mass have if the string is not to break?

Homework Equations

F=m*a
T=r*F*sin(theta)
v=r*omega

The Attempt at a Solution

I have absolutely no idea where to even begin with this. My idea was to find the acceleration when the force is 450 N, but that yields the linear acceleration not the angular one which does nothing for me. I am stumped.

 

[Oddbio]

Well even with the object being swung around at a constant (angular) speed you will still have a force. Because the direction of the velocity vector is changing, and acceleration is change in velocity. So you are looking for the acceleration inward, which is the centripetal acceleration, the force vector would be pointing from the object toward the center of the circle as it spins.

So the equations you are missing are:
$$a_{centripetal}=\frac{v^{2}}{r}$$
OR equivalently:
$$a_{centripetal}=\omega^{2}r$$

 

[kerimek]

This is a classic example of circular motion, where the centripetal force (provided by the tension in the string) keeps the mass moving in a circular path. To find the maximum speed that the mass can have without breaking the string, we can use the formula for centripetal force:

F = m*v^2 / r

Where F is the force (in this case, the maximum tension of 450 N), m is the mass (0.40 kg), v is the velocity, and r is the radius of the circular path (0.75 m).

Rearranging the equation, we get:

v = √(F*r/m)

Substituting in the known values, we get:

v = √(450 N * 0.75 m / 0.40 kg) = 15.6 m/s

Therefore, the maximum speed that the mass can have without breaking the string is 15.6 m/s. Any faster and the tension in the string would exceed its maximum capacity and it would break.