A small remote-controlled car with mass $1.60$ kg moves at a constant speed of $v = 12.0$ m/s in a track formed by a vertical circle inside a hollow metal cylinder that has a radius of $5.00$ m (Fig. E $5.45$ ). What is the magnitude of the normal force exerted on the car by the walls of the cylinder at point $B$ (top of the track)?

Question

A small remote-controlled car with mass $1.60$ kg moves at a constant speed of $v = 12.0$ m/s in a track formed by a vertical circle inside a hollow metal cylinder that has a radius of $5.00$ m (Fig. E $5.45$ ). What is the magnitude of the normal force exerted on the car by the walls of the cylinder at point $B$ (top of the track)?

Diagram of a remote-controlled car on a vertical circular track, showing points A and B.

Accepted Answer

Welcome back everybody. We are making observations about a ferris wheel. So this circle will represent our ferris wheel. And at the very top we have a cart that holds a child, a seat car that holds a child and we are told a couple different things about the system. We are told that the mass of the kid is 50 kg. We are told that the radius of the ferris wheel is 15 m and that it has a period of 20 seconds. And we are tasked with finding what the normal force the normal force is going to be of the seat acting on the child. Well, in order to figure that out, we want to draw out a couple other different forces that are at play here. Of course we have the force due to gravity, but we also have the centrifugal force. The centrifugal force. Now, centrifugal forces like the centripetal force. That's just except it's acting in the opposite direction here. Right? So we have M. V. Squared over R. Now, just by convention, I'm actually going to deem the downward direction, our positive Y direction, making both the normal force and the centrifugal force negative. So now, let's use Newton's second law here. Newton's second law states that the sum of all forces in a given direction is equal to some mass times some acceleration. Well, we have M G minus Mv squared over R minus r. Normal force is equal to what? Well the kid stays in his seat. So he doesn't have an acceleration, right? He is therefore the right side of this equation is just going to be zero. If we add the normal force to both sides here, we then get that. The normal force is equal to MG minus mv squared over R. But we so we have all these terms except for this ve term right here, what is ve equivalent to we have a formula this we have that V is equal to two pi times the radius divided by our period. So let's go ahead and plug in our values. Here we have two pi times 15, divided by 20 gives us a velocity of 4. m per second. Now we are ready to find our normal force. So we have 50 times the acceleration due to gravity minus 50 times 4.7 squared divided by 15. This gives us a normal force of newtons corresponding to our answer choice of B. Thank you all so much for watching. Hope this video helped. We will see you all in the next one.

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Key Concepts

Centripetal Force

Normal Force

Gravitational Force