Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) on a horizontal frictionless surface with a tangential speed of 5 m/s. the string has been slowly winding around a vertical rod, and a few seconds later the length of the string has shortened to 0.250 m. what is the instantaneous speed of the mass at the moment the string reaches a length of 0.250 m?

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum is

where is the position and the linear momentum.

We also know that the torque is

but, as the linear momentum is this means that is parallel to the velocity, and the first term must equal zero

so

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

this means, for the angular momentum measure from the rod:

that means :

So, the magnitude of initial angular momentum is :

but the angle is 90°, so:

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

For our final angular momentum we have:

and the radius is 0.250 m and the mass is 2.00 kg

but, as the angular momentum is constant, this must be equal to the initial angular momentum

b

step-by-step explanation: a third of 24 is 8, 8 times 3 is 24 so 8 - 2 would be 6

answer; equivalent to a one(1) horsepower motor;