Rolling Cylinders

  • #1

Homework Statement



A thin cylindrical shell and a solid cylinder have the same mass and radius. The two are released side by side and roll down, without slipping, from the top of an inclined plane that is 1 m above the ground. The acceleration of gravity is 9.8 m/s2.

Find the final linear velocity of the (b) thin and (a) solid cylinder.

(c) When the first object reaches the bottom, what is the height above the ground of the other object?

Homework Equations



E = mgh + 1/2mv^2 + 1/2Iw^2

w = v/r

I = mr^2

I = 1/2mr^2

The Attempt at a Solution



(a) vf (solid) = (4gh/3)^1/2 = 3.61 m/s

(b) vf (thin) = (gh)^1/2 = 3.13 m/s

(c)???

I don't know where to start for part c.
 

Answers and Replies

  • #2
berkeman
Mentor
61,348
11,870
Solve for the time it takes for the first cylinder to reach the ground, and plug that time back into the equation of motion of the other cylinder to find its height at that time.
 
  • #3
I still don't know which equation to use!

Is it:

v = -1/2gt
t = 0.368

and then what?

or am I still off?
 
  • #4
LowlyPion
Homework Helper
3,097
5
I still don't know which equation to use!

Is it:
v = -1/2gt
t = 0.368

and then what?
or am I still off?

Consider

y = 1/2*a*t2
 
  • #5
well then:

y = (1/2)(9.8)(0.368)^2
y = 0.66 m

BUT this answer is wrong!

The answer is 0.25 m.

Am I using the right acceleration and time?
 
  • #6
LowlyPion
Homework Helper
3,097
5
well then:

y = (1/2)(9.8)(0.368)^2
y = 0.66 m

BUT this answer is wrong!

The answer is 0.25 m.

Am I using the right acceleration and time?

Don't you want to consider the time of the faster, in the equation of the distance the slower will have gone and then take the difference?
 
  • #7
Don't you want to consider the time of the faster, in the equation of the distance the slower will have gone and then take the difference?

I'm sorry but what does this mean?

I don't understand what you just said.
 
  • #8
SOME BODY PLEASE HELP!!! I'm so confused on this problem!
 
  • #9
LowlyPion
Homework Helper
3,097
5

The Attempt at a Solution


(a) vf (solid) = (4gh/3)^1/2 = 3.61 m/s
(b) vf (thin) = (gh)^1/2 = 3.13 m/s
(c)???

You've found that

v_solid2 = 4/3*gh
v_thin2 = gh

Consider also then that

v2 = 2*a*x

If you explore the relationship of the ratio of the velocity2 one to the other you will have a ratio of the accelerations don't you?

Armed with that you also know that

x = 1/2*a*t2

What happens then when you plug in the acceleration of the slower, to the equation of the faster? For the same t2 what will the drop have been?

All you need to do then is determine how much further the slower has to go ... the answer to part C.
 

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