Circular motion of rope and ball

[parabol]

Homework Statement

A motor is used to rotate a ball attached to a rope in a vertical plane. The mass of the ball is 2kg and the length of the rope is 3m. Ignoroing air resistance and the mass of the rope, calculate:

a) The minimum motor speed in rpm that will maintain the ball in a circular motion

b) The total energy of the ball at the top of the circle

Homework Equations

a

The Attempt at a Solution

a) If the motor is providing enough speed to maintain a circular motin then at the balls highest point all hte centripetal accelration is provided by gravity.

a = g = 9.81 = w²r= w² x 3

so

$$\omega=\sqrt{\frac{9.81}{3}} = 1.808 rad/s$$ rad/s

$$\omega=\frac{1.808}{2\Pi}=0.288rev/s$$

$$\omega=0.288rev/s . 60 = 17.28 rpm$$


b)

v = rw = 3 x 1.808 = 5.424 m/s

Potential Energy at the top of the circle = mgh = 2 x 9.81 x 3 = 58.86 J
Kinetic Energy = 1/2 m v^2 = .5 x 2 x 5.424 = 29.42 J

Total Energy at the top of the circle = 58.86 + 29.42 = 88.28 J



Hi, just after sanity check and to make sure the assumptions I have made in part a) are correct. I'm not 100% on them.

Thanks in advance

Parabol

 

[Redbelly98]

(a) looks good.

(b) is probably okay, but the question is ill-posed. They do not say where to take the potential energy to be zero. If we take h=0 at the center of the circle, then you're answer is correct.