Rotational Motion Homework: Hoop Rolling on 15 Deg Incline

[pb23me]

Homework Statement

A hoop rolling on a horizontal surface with a speed v=3.3m/s when it reaches a 15 deg incline.How far up the incline does it go?How long will it be on the incline before it reaches the botttom?

Homework Equations

I=mr²
conservation energy going up=2gy_f=r²+v_i²
conservation energy going down=2gy_i=r\omega²+v²
y_f=sin(15)
circumference=2r\pi=8.5m

The Attempt at a Solution

i used the conservation of energy equation and solved for r getting r=1.36m. I had my calculator in radian mode not sure what mode i should have had it in. Then i plugged the radius back in and got y_f=.65m , sin(15) was also .65m so hypotenuse =1m. Then i used conservation of energy down equation and solved for [iex]\omega[/itex] getting \omega=2rad/s i found how many radians the hoop went through by noting that the hypotenuse was 1m long so 1/8.5(2\pi)=.74 rad
so \omega=.74rad/t=.37s

 

[Doc Al]

I=mr²

Good.

conservation energy going up=2gy_f=r²+v_i²

Rethink this. You're missing an ω². (Check the units of each term.)

3. The Attempt at a Solution [/b] i used the conservation of energy equation and solved for r getting r=1.36m.

The radius is not given and could be anything. (That should tip you off that something's wrong with your conservation equation.)

Hint: The hoop is rolling without slipping, so how are v and ω related?

 

[pb23me]

oh ok i forgot to put the\omega in so it should be (r\omega)²+v²=2gy_f so y_f=1.1m and distance up =1.7m I am confused about what mode to have my calculator in? rads or degs?

 

[Doc Al]

oh ok i forgot to put the\omega in so it should be (r\omega)²+v²=2gy_f so y_f=1.1m

Good.

and distance up =1.7m I am confused about what mode to have my calculator in? rads or degs?

Since the angle is given in degrees, use the degree mode.

 

[pb23me]

oh...that makes sense haha

 

[pb23me]

ok so i used the equation 2gyf=r^2+vi^2 solved for yf and got yf=1.1m so sin(15)=1.1/h hypotenuse=.43m I am getting stuck on the last part now,solving for the time.I think i need the radius of the loop but can't seem to figure it out...

 

[Doc Al]

ok so i used the equation 2gyf=r^2+vi^2 solved for yf and got yf=1.1m so sin(15)=1.1/h hypotenuse=.43m

Careful with your arithmetic.

I am getting stuck on the last part now,solving for the time.I think i need the radius of the loop but can't seem to figure it out...

You won't be able to figure out the radius, but luckily its irrelevant. Hint: What's average speed of the hoop as it goes up (or down) the ramp?

 

[pb23me]

Ok hypotenuse =4.25m and I thought the speed going up would be different than the speed going down because the hoop has an initial velocity going up.

 

[pb23me]

Nevermind that the speeds are the same