Inertia, frictionless pulley, and a weight

In summary, the problem involves a frictionless pulley system with a uniform solid disk of mass 2.5 kg and radius 0.2 m. A 1.5 kg stone is attached to a light wire wrapped around the pulley and the system is released from rest. The first part of the question asks for the distance the stone must fall for the pulley to have 4.5 J of kinetic energy. The second part asks for the percentage of total kinetic energy that the pulley has. The solution involves using the equation for mechanical energy and the formula for moment of inertia to find the final velocity of the pulley and then using conservation of energy to solve for the distance the stone must fall.
  • #1
jaredmt
121
0

Homework Statement


A frictionless pulley has the shape of a uniform sold disk of mass 2.5 kg and radius .2m. A 1.5kg stone is attached to a very light wire that is wrapped around the rim of the pulley, and the system is released from rest. (a) How far must the stone fall so that the pulley has 4.5J of kinetic energy? (b) What percent of the total kinetic energy does the pulley have?

Homework Equations


I for solid disk: I = (1/2)(MR^2)

The Attempt at a Solution



first i plugged in this equation:
K = (1/2)IW^2 + (1/2)MV^2
i replaced W with V/r
K = (1/2)I(V/r)^2 + (1/2)MV^2
first i had to find I:
using formula above: I = (1/2)(2.5)(.2^2) = .05 kgm^2
then i plugged in everything to find final velocity :
4.5J = (1/2)(.05kgm^2)(V/.2m)^2 + (1/2)(1.5kg)V^2
then i solved for V and got V = 1.8 m/s

ok now I am not entirely sure what to do... i need the distance that the rock drops. but to do that i need to find the acceleration.
radial acceleration is V^2/r = 16.2 rad/s^2

then i tried V^2 = Vo^2 + 2(a)(X-Xo)
Vo=0 and Xo=0 so it becomes:
V^2 = 2(a)(x)

i have value for V. i know I am supposed to be able to figure out A so i can solve for x. but I am not sure if I am converting from radial acceleration to regular acceleration correctly because i keep getting the wrong answer.

i thought this was the formula for that:
a^2 = Arad^2 + Atan^2
and Atan = Arad(r) right? because I am doing something wrong... idk what it is
the answer is supposed to be 67.3 cm
 
Last edited:
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  • #2
jaredmt said:
first i plugged in this equation:
K = (1/2)IW^2 + (1/2)MV^2
i replaced W with V/r
K = (1/2)I(V/r)^2 + (1/2)MV^2
first i had to find I:
using formula above: I = (1/2)(2.5)(.2^2) = .05 kgm^2
then i plugged in everything to find final velocity :
4.5J = (1/2)(.05kgm^2)(V/.2m)^2 + (1/2)(1.5kg)V^2
then i solved for V and got V = 1.8 m/s
4.5J is the KE of the pulley, not the total KE.

Figure out the total KE. Then ask where that energy came from. (Hint: Conservation of mechanical energy.)
 
  • #3
thanks for the help. i got part (b) which is Kpulley/Ktotal = 4.5/9.89 x 100 = 45.5% :)

but I am still stuck on part (a). since i got the second part right i know that my velocity is right now. the W = 13.4 rad/s and V = Wr = 13.4(.2) = 2.68 m/s.
this time i tried solving the distance by finding the change in the angle:
Arad = (W^2)(r) = 36 rad/s^2
W^2 = 2(Arad)(feta)
13.4^2 = 2(36)(feta)
feta = 2.494 rads
this doesn't come out to 67.3m :(

edit: SWEET! i figured it out cus of ur mechanical energy hint!
i did:
Ktotal = mgh
9.89 = 1.5(9.8)h
h = .673m = 67.3 cm

thanks a lot!
 
Last edited:
  • #4
I can't quite follow what you're doing. In any case, instead of messing around with speeds, use conservation of energy. Compare the mechancial energy of the system before and after the stone falls.
 

1. What is inertia?

Inertia is the tendency of an object to resist changes in its state of motion. This means that an object at rest will remain at rest and an object in motion will continue moving in the same direction and at the same speed unless acted upon by an external force.

2. How does friction affect inertia?

Friction is a force that opposes the motion of an object. In some cases, friction can decrease the inertia of an object by slowing it down. For example, when a car is in motion, the friction between the tires and the road causes the car to eventually come to a stop.

3. What is a frictionless pulley?

A frictionless pulley is a pulley with no friction or resistance. This means that when a rope or belt is looped over the pulley, it can move freely without any loss of energy due to friction. In other words, the tension in the rope will be the same on both sides of the pulley, making it easier to lift heavy objects.

4. How does a weight affect a frictionless pulley?

In a frictionless pulley system, the weight or force acting on one side of the pulley will be equal to the weight or force on the other side. This means that a weight on one side of the pulley can be balanced by a smaller weight on the other side, making it easier to lift heavy objects.

5. What is the relationship between inertia and a weight on a frictionless pulley?

Inertia and weight are directly proportional in a frictionless pulley system. This means that the greater the weight on one side of the pulley, the greater the inertia of the system, and the more force is required to move it. Similarly, a lighter weight will have less inertia and require less force to move.

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