A solid rubber ball rests on the floor of a railroad car when ...

Question

A solid rubber ball rests on the floor of a railroad car when the car begins moving with acceleration a. Assuming the ball rolls without slipping, what is its acceleration relative to the car?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. What is the acceleration of a solid rubber weight resting on the train floor relative to the car? If the train accelerates with an acceleration, a assuming the object rolls without slipping. So that's our goal. Our angles, we're ultimately trying to figure out what the acceleration is of this rubber or I should be specific this solid rubber weight that's resting on the train floor inside of this train car relative to the train. If the car, the train car itself accelerates with an acceleration of A and we're also assuming that the object rolls without slippage. Awesome. So our final answer is we're trying to figure out what this acceleration value is of this solid rubber weight. Awesome. So we're also given some multiple choice answers. Let us note that all of our multiple choice answers are some number multiplied by a subscript train relative to the ground. So it's the acceleration of the train relative to the ground So A would be negative 5/7. So negative 5/7 B is zero C is 1/7 and D is 4/8. Awesome. Moving right along here. First off, let us draw a diagram to help us better visualize all the forces at work in this particular problem. So I've gone ahead and come and put together a fancy little diagram for us to look at here. So the circle represents our solid rubber weight that's resting on the train floor. And as we can see in the vertical direction is the Y direction as shown in our little xy coordinate plane up here. So the vertical directions in the Y direction and the horizontal direction is in the X direction. So going to the left, we have our mass of the train multiplied by the vector a acceleration of the train going down as we have the weight force, which is the mass of the. So I said mass of the church. So we have the mass of the ball and we also, so we have the mass of the ball, I should say. So it's the mass of the ball. So that's capital M not the mass of the train. And then we have capital M once again, for the mass of the ball multiplied by gravity because we have the mass of the ball multiplied by gravity which is gravity and the mass of the ball are pulling it back towards the train floor trying to prevent it from floating off into outer space. And then we have the radius capital R, which is like the halfway point of our little circle here. And we also of course have the normal force pointing in the upwards direction. And then since the ball wants to roll to the left, we have our friction force which is vector FFR. So our friction force which is a vector going to the right. So if our ball is rolling to the left, the friction force is rolling to the right. OK. Now that we have a visualization of what's going on in this problem, we now need to note that in the accelerating reference frame, the f fictitious force. So the friction force of the magnitude capital M multiplied by a train, relatives of the ground must be added in the opposite direction to the acceleration of the train as this is shown in our diagram. And since the weight rolls without slipping, we can recall and use the following equation for the angular acceleration. Let us note that the angular acceleration is denoted by alpha and alpha is equal to a which is the weight of the train relative to the GR. So it's the acceleration, a subscript weight relative to train. So I'm just gonna call it W RT. So weight relative to train all divided by capital R which is our radius. And let us call this equation one. And we need to note once again that alpha is the angular acceleration A which is the weight relative to the train is equal to the linear acceleration of the weight relative to the train. And capital R is the radius of the weight. So now we need to consider the clockwise torque to be positive. And if we recall and use Newton's second law, we can write that the sun of torque, which we're gonna denote Tau, the Greek symbol, Tau to be torque. So the sum of torque is equal to negative, the friction force multiplied by R which is all equal to the moment of inertia multiplied by alpha. And let's call this equation two. And once again, Tau is torque, F subscript fr is the fictious force and I is the moment of inertia. And let's quickly note here in blue here that the moment of inertia I, the equation is equal to since this is a solid rubber weight. So I'm gonna assume that this is a solid rubber weight, it appears to be a sphere. So we're gonna use the sphere moment of inertia equation. So it's gonna be 2/5 multiplied by capital M multiplied by capital R squared. Awesome. So now if we substitute the value of I, which once again, I is the moment of inertia into equation two and plug in the value of alpha from equation one into equation two, we will get negative FFR. So the negative F friction force multiplied by, you need to multiply this by R which is equal to 2/5 multiplied by M multiplied by capital R squared. Awesome. And then we need to take this and we need to multiply all of that by the acceleration which is a, so a subscript weight relative to the train divided by capital R. And let us call this equation three. Awesome. Let us also note that when we simplify, we will get that F subscript fr is equal to negative 2/5 multiplied by M multiplied by a weight relative to train. So now we need to recall and write the equation for the force. So let us recall that the sum of FX is equal to FFR. So the friction force minus M multiplied by A and this time it's acceleration subscript train relative to ground. So I'm gonna to simplify it to TRG. So train relative to ground is all equal to M multiplied by a subscript weight relative to train. And let us call this equation four. Nice. So we're a train relative to wait. So train relative to wait in this case is going to be the acceleration of the train relative to the ground. Awesome. Awesome. Awesome. Awesome. OK. So now we need to note that a subscribed weight relative to train is the acceleration of the weight relative to the train. So now we need to substitute equations 32 equation four and we will get that negative 2/5 multiplied by M multiplied by A and this is a subscript weight relative to train minus M multiplied by a, a subscript train relative to ground, which is equal to M multiplied by a subscript and make sure we're writing this neat. So it's a subscript weight relative to train. Awesome. So, moving right along here now, if we rearrange this equation to isolate and solve four, because our end goals, we're trying to find a subscript weight relative to train, We will get that A is or better yet. Don't forget because we need to make sure we're solving for specifically a subscript weight relative to train is equal to negative 5/7 multiplied by a subscript train relative to ground. And that's it we've solved for our final answer. Hooray. So looking at our multiple choice answers, the correct answer has to be the letter A negative 5/7 multiplied by a subscript train relative to ground. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

A solid rubber ball rests on the floor of a railroad car when the car begins moving with acceleration a. Assuming the ball rolls without slipping, what is its acceleration relative to the car?

Key Concepts

Newton's Second Law of Motion

Rolling Motion

Relative Acceleration

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