How can the time it takes for a hoop to roll up an incline be calculated?

[gillyr2]

Homework Statement

A hollow cylinder (hoop) is rolling on a horizontal surface at speed v = 3.8 m/s when it reaches a 14^\circ incline. how far up the incline will it go

Homework Equations

mgH = .5mv^2 + .5Iw^2
v=sqrt(10/7 *g * H)

The Attempt at a Solution

i need to solve for time, the distance is equal to 6.1 m

 

[Doc Al]

Homework Equations

mgH = .5mv^2 + .5Iw^2

This is the one you need.

v=sqrt(10/7 *g * H)

Where does this come from?

 

[gillyr2]

i canceled the m's and solved for v. where .5Iw^2 = 1/5 * Mv^2

i know the height = sin(14)*distance

 

[Doc Al]

where .5Iw^2 = 1/5 * Mv^2

Why do you think this? What's I?

 

[gillyr2]

I = 2/5 M R^2
w = v/R

.5 (2/5 M R^2) (v/R)^2 = .5 Iw^2

 

[gillyr2]

oh ok i figured it out i was doing the sphere instead of hoop.
for the time can we use x = vt + .5gt^2 ?

 

[gillyr2]

v=gsin(14)t

how exactly are we suppose to find the time without the decelleration?

 

[JTHERO]

Try

a=(2/3)*g*sin(14)

 

[merryjman]

When you are solving the problem using energy, which is what you're doing, there is no need to look at time. Why do you need the time? Aren't you looking for the distance up the ramp?

 

[gillyr2]

i already found the distance up the ramp, but the next part is to find the time is took to get up and down the ramp and I am not sure how.

 

[merryjman]

Oh, OK. Try thinking about it as a rotational kinematics problem. So you want to find [tex]\alpha[/tex], not a.

 

[gillyr2]

yes i have been looking for alpha but I am not sure how.

 

[gillyr2]

am i suppose to use alpha = w^2/2pheta

 

[gillyr2]

i can't get it. can someone help please/

i have

w = v/R

and

alpha = w^2/2pheta

 

[kimlovesyou24]

.5(m)(v^2) + (.5)(m)(v^2) = mgh
therefore: (.5)(3.8^2) + (.5)(3.8^2) = (9.8)(h)
h ends up being 1.5 and when you divide it by sin(14) to get the hypotenuse, the answer is 6.1 up the ramp.

i don't know about the second part

 

[kimlovesyou24]

.5(m)(v^2) + (.5)(m)(v^2) = mgh
(.5)(3.8^2) + (.5)(3.8^2) = (9.8)(h)
h ends up being 1.5 and when you divide it by sin(14) to get the hypotenuse, the answer is 6.1m up the ramp.

 

[merryjman]

Make it a torque problem. What torque(s) are slowing down the hoop? What is a good axis of rotation to choose? (Hint - don't choose the center of the hoop as your axis)

 

[Doc Al]

i already found the distance up the ramp, but the next part is to find the time is took to get up and down the ramp and I am not sure how.

You have the initial and final speeds and the distance (from the first part). To find the time, just use: Distance = average speed X time. What's the average speed?