Circular motion with car on hill

[tomboi03]

Homework Statement

A toy car with mass m=15.0g rolls without friction along a track that goes over a hill. The hill is in the shape of an arc of radius R=35.0cm. The center-of-mass (CM) of toy car is 0.5cm beyond the point of contact with the track. At the instant shown in the drawing its speed is 0.900m/s in the direction shown. Draw an accurate freepbody diagram of the toy car in the box and find all the following vector quantities indicated (both magnitude and direction). Clearly specify direction with respect to any unambiguous axis direction and make clear your choice of coordinate system. Show your work.

Homework Equations

The relevant equations are used in my attempt to solve this problem

The Attempt at a Solution

a. F_net= F_tangential +F_G= ma_T+mg
b. a= F_net/m= (ma_T+mg)/m = a_T+g
c. N_T,C(Normal force by the track on the car)= -F_centripetal=-mgsin30= -(0.015kg)(9.8m/s²)sin30
d. F_centripetal=mgsin30=(0.015kg)(9.8m/s²)sin30
e. F_tangential=ma_T= (0.015kg)((0.800m/s²)²/0.35m)
f. a_centripetal=9.8m/s^2 (pointing toward center or circle)
g. a_tangential= v_T²/r= (0.800m/s²)²/0.35m

I don't know if this is all correct.. but if it is... i have the free body diagram... if it isn't... then i don't have the free body diagram and I don't understand what the F_netwould be... is there any other force on the car that i should know about?

Thanks

 

[anathema]

for posting this problem and attempting to solve it. Let me help clarify some things and correct some errors in your solution.

Firstly, your free body diagram should include all the forces acting on the toy car. These include the normal force from the track, the weight of the car, and the centripetal force due to its circular motion on the track. The normal force and the weight should be shown acting on their respective points of application, while the centripetal force should be shown acting towards the center of the circular path.

Secondly, your equations are mostly correct, but you have some errors in your calculations. For example, in part c, the normal force should be equal to the centripetal force, not the other way around. Also, in part e, the tangential acceleration should be calculated using the radius of the circular path, not the height of the hill.

Using these corrections, your solution should look something like this:

a. Fnet= NT + mg + FC= ma
b. a= Fnet/m= (NT + mg + FC)/m = aT + g + aC
c. NT= FC= mgsin30= (0.015kg)(9.8m/s^2)sin30
d. FC=mgsin30=(0.015kg)(9.8m/s^2)sin30
e. aT= v^2/r= (0.900m/s)^2/0.35m
f. aC= v^2/r= (0.900m/s)^2/0.35m
g. The direction of aT and aC should be tangential to the circular path, which is in the direction shown in the drawing.

I hope this helps. Keep in mind that the direction of the forces and accelerations may change as the car moves along the path, so it's important to carefully consider the situation at each point. Good luck with your studies!

 

[nlightner]

for your question. It looks like you have made a good attempt at solving this problem. Let's go through it step by step and see if we can clarify any confusing points.

First, the problem states that the car is rolling without friction, which means that there is no frictional force acting on the car. This means that the only forces acting on the car are the normal force from the track and the force of gravity.

Next, you correctly identified that the net force on the car is equal to the sum of the tangential force and the force of gravity. This is because the car is moving in a circular motion, so there must be a centripetal force acting on it to keep it on the track. This force is provided by the normal force from the track.

Your calculation of the normal force (NT,C) is correct, but it is important to note that the angle of the hill is not given in the problem. You assumed that it was 30 degrees, but this information is not given. This means that you cannot calculate the tangential force (Ftangential) or the centripetal force (Fcentripetal) because these depend on the angle of the hill. So, we cannot calculate the value or direction of the net force on the car.

Your calculations for the tangential acceleration (atangential) and the centripetal acceleration (acentripetal) are correct. These are the magnitudes of the accelerations in the tangential and centripetal directions, respectively. The direction of the centripetal acceleration is always towards the center of the circle, while the direction of the tangential acceleration depends on the direction of the velocity at that point.

Finally, it is not clear what you mean by "Fnet" in your question. If you are asking about the net force on the car, as stated earlier, we cannot calculate it without knowing the angle of the hill. If you are asking about the net force acting on the car, then it is simply the force of gravity since there is no friction.

In terms of the free body diagram, it should show the forces acting on the car at the instant shown in the drawing. These would be the normal force (NT,C) and the force of gravity (FG). The direction of the normal force should be perpendicular to the track and the direction of the force of gravity should be downwards.

I hope this helps clarify the problem. Remember, when solving physics problems, it is