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.

 

[Mene]

oid


Your calculations and assumptions seem to be correct. In circular motion, the centripetal acceleration is always provided by an external force, in this case gravity. So at the top of the circle, the only force acting on the ball is gravity, and it provides all the centripetal acceleration needed to maintain the circular motion. Your calculation for the minimum motor speed to maintain the circular motion is also correct. In part b), you have correctly calculated the total energy of the ball at the top of the circle by considering both its potential and kinetic energy. Good job!