# Find Speed of Cylindrical Shell on Incline-Loop

• chrisfnet
In summary: Then, I can use conservation of energy to find the velocity at the top of the loop by equating the potential energy at the top of the loop to the sum of the kinetic energy at the bottom of the incline and the rotational kinetic energy at the top of the loop. In summary, the problem involves a cylindrical shell rolling down an incline without slipping. The goal is to find the velocity of the shell at the top of the loop. Using conservation of energy, we can first find the velocity at the bottom of the incline using the equation vf = sqrt(gh). Then, we can use this velocity and the conservation of energy equation to find the velocity at the top of the loop by equating the potential energy at the top of
chrisfnet

## Homework Statement

http://img688.imageshack.us/img688/5277/loopb.jpg

A cylindrical shell is released from rest and rolls down an incline without slipping. Find speed of shell at the top of the loop.

h(incline) = 2.0m
h(loop) = 1.3m

## The Attempt at a Solution

vf = sqrt(g * yi)
vf = sqrt(9.8m/s^2 * 2.0m)
vf = 4.42719m/s = vi(loop)

KEi = PEf + KEf
1/2mvi^2 = mgy + 1/2mvf^2 + 1/2mr^2(v^2/r^2)
1/2vi^2 = gy + 1/2vf^2 + 1/2vf^2
1/2(4.42719m/s)^2 = -9.8m/s^2(1.3m) + 1/2vf^2 + 1/2vf^2
vf = 1.71464m/s

Would this be correct?

Last edited by a moderator:
No. First, it would be $$v_{f}=\sqrt{2gh}$$ but you can't use that equation anyway because in addition to translational kinetic energy, there is rotational kinetic energy.

Start off by looking up or deriving the moment of inertia of the cylinder for the rotation. Use your conservation of energy equation with change in potential energy on one side and change in rotational and translational kinetic energy on the other. Because the cylinder is rolling without slipping, you can find the angular velocity in terms of the linear velocity. Now you can solve for linear velocity.

Actually, sqrt(2gh) would be right if it were a block going down the incline. However, it's not. Since it's a cylindrical shell, the equation sqrt(gh) is correct.

I'd use:

mgh = 1/2mvf^2 + 1/2Iwf^2

Which simplifies to vf = sqrt(gh) after substitution.

This would give me the velocity at the bottom of the incline.

## 1. How do you calculate the speed of a cylindrical shell on an incline-loop?

To calculate the speed of a cylindrical shell on an incline-loop, you will need to use the formula v = √(gRsinθ), where v is the speed, g is the acceleration due to gravity (9.8 m/s²), R is the radius of the loop, and θ is the angle of the incline. This formula assumes that there is no friction or air resistance present.

## 2. What factors affect the speed of a cylindrical shell on an incline-loop?

The speed of a cylindrical shell on an incline-loop is affected by several factors, including the angle of the incline, the radius of the loop, the acceleration due to gravity, and any external forces like friction or air resistance. In general, a steeper incline or smaller radius will result in a higher speed, while friction and air resistance will decrease the speed.

## 3. How does the mass of the cylindrical shell impact its speed on an incline-loop?

The mass of the cylindrical shell does not directly impact its speed on an incline-loop. However, a heavier shell may require more force to maintain its speed, which could be affected by factors such as friction and air resistance. In general, a heavier shell will have a slower speed than a lighter shell, assuming all other factors are constant.

## 4. What is the difference between the speed of a cylindrical shell on a flat incline and an incline-loop?

The speed of a cylindrical shell on a flat incline will be constant, as there is no change in direction. However, on an incline-loop, the speed will vary as the shell moves through the loop and experiences changes in direction. Additionally, the speed on an incline-loop will be affected by the radius of the loop, which is not a factor on a flat incline.

## 5. Can the speed of a cylindrical shell on an incline-loop be greater than the speed of an object in free fall?

Yes, it is possible for the speed of a cylindrical shell on an incline-loop to be greater than the speed of an object in free fall. This can occur if the incline-loop has a steep angle and a small radius, resulting in a higher speed for the shell. However, this also depends on the mass of the objects and any external forces present.

Replies
5
Views
2K
Replies
6
Views
2K
Replies
4
Views
1K
Replies
12
Views
11K
Replies
10
Views
2K
Replies
13
Views
2K
Replies
1
Views
3K
Replies
2
Views
2K
Replies
23
Views
2K
Replies
19
Views
2K