A potato of mass M is attached to a massless rope of length L.
The rope is used to move the tater in a vertical circle. At the bottom of the circular path, the potato almost skims the ground. At the top of the circular path, the tension in the rope is twice the weight of the potato. Air resistance is negligible. Write all answers in terms of M, L, and g.
a. Determine the magnitude and direction of the net force acting on the potato at the top of the circle.
b. Determine the speed, v, of the potato at the top of the circle.
You decided to cut the rope when the potato is at the top of the circular path.
c. How long does it take the potato to reach ground?
d. How far away, horizontally, should you place a trash can so that it catches the potato (assume the trash can has minimal height, that way your potato won’t hit it from the side)
a. The tension from the string acting on the mass is Tt = 2mg, acting downwards, towards the centre of the circle.
The weight of the potato is also acting upon the mass. This is a force of mg, acting vertically downwards, towards the centre of the circle.
Net force, acting downwards is N = Tt + mg = 2mg + mg = 3mg
N = 3mg
b. Centripetal force is given by Fc = mv^2/r, where r = L
This is the force acting outwards, away from the centre of the circle.
There is a force of mg acting inwards, from the weight of the potato.
The tension in the rope is Tt, given by Tt = Fc – mg. Substituting ...
2mg = mv^2/L – mg
3mg = mv^2/L
3gL = v^2
v = √(3gL)
c. The potato is, in effect, falling through a vertical distance of h = 2L, with zero initial vertical velocity.
The equation of motion for such movement is given by
s = ut + (1/2)at^2, where s = h. Substituting ...
2L = 0 + (1/2)gt^2
t^2 = 4L/g
t = 2√(L/g)
d. The horizontal velocity of the potato is the tangential velocity of the potato when it reaches the top of the circle, which is the speed given in part b. i.e.
v = √(3gL)
The potato moves at this velocity for the time given in part c. i.e.
t = 2√(L/g)
Distance travelled is d = vt,
d = √(3gL) * 2√(L/g) = 2√(3L^2)
d = 2√3.L
