Torque Calculation for Jacked-Up Car Changing Tires

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Can anyone help me with these problems? Thank-you

1.) In a circus performance, a large 4.0 kg hoop with a radius of 1.8 m rolls without slipping. If the hoop is given an angular speed of 5.0 rad/s while rolling on the horizontal and is allowed to roll up a ramp inclined at 12° with the horizontal, how far (measured along the incline) does the hoop roll?

I know I have to use the total mechanical energy formula to solve for the height, then find the sin of 12°. I tried canceling out, and not but each answer I got was wrong. So It would be 1/2mv²+1/2mr²(v²/r²)=mgh, which is 1/2(4)(5²)+1/2(4)(1.8²)(5²/1.8²)=(4)(9.8) when I canceled out I got 50=39.2h (h=1.28), when I did not I got 100=39.2h (h=2.55). After that, I solved for H, with sin12=oppositite/hypotenuse, so I got H=-2.38 and -4.75. I know that distance can't be negative, so I'm pretty sure my signs aren't my problem.
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2.) A mechanic jacks up a car to an angle of 8.0° to change the front tires. The car is 3.20 m long and has a mass of 1160 kg. Its center of mass is located 1.12 m from the front end. The rear wheels are 0.40 m from the back end. Calculate the torque exerted by the car around the back wheels.

I know it's going to be negative, and that Torque= Fr sin theta
 
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hshphyss said:
1.) In a circus performance, a large 4.0 kg hoop with a radius of 1.8 m rolls without slipping. If the hoop is given an angular speed of 5.0 rad/s while rolling on the horizontal and is allowed to roll up a ramp inclined at 12° with the horizontal, how far (measured along the incline) does the hoop roll?

I know I have to use the total mechanical energy formula to solve for the height, then find the sin of 12°. I tried canceling out, and not but each answer I got was wrong. So It would be 1/2mv²+1/2mr²(v²/r²)=mgh, which is 1/2(4)(5²)+1/2(4)(1.8²)(5²/1.8²)=(4)(9.8) when I canceled out I got 50=39.2h (h=1.28), when I did not I got 100=39.2h (h=2.55). After that, I solved for H, with sin12=oppositite/hypotenuse, so I got H=-2.38 and -4.75. I know that distance can't be negative, so I'm pretty sure my signs aren't my problem.

Two problems: (1) you are mixing up angular speed with linear speed (5.0 rad/s is not v!), and (2) realize that "h" is the change in height of the center of mass of the hoop and that when the hoop is on the incline its center is not directly above the point of contact (the initial height of that center of mass is not at zero).

2.) A mechanic jacks up a car to an angle of 8.0° to change the front tires. The car is 3.20 m long and has a mass of 1160 kg. Its center of mass is located 1.12 m from the front end. The rear wheels are 0.40 m from the back end. Calculate the torque exerted by the car around the back wheels.

I know it's going to be negative, and that Torque= Fr sin theta
The force creating the torque is the car's weight: Where does it act? Also realize that "theta" is the angle between "r" and "F". (Draw yourself a diagram.)