How is the angular velocity ωᵣ of the rear wheel of a bicycle ...

Question

How is the angular velocity ωᵣ of the rear wheel of a bicycle related to the angular velocity ωբ of the front sprocket and pedals? Let Nբ and Nᵣ be the number of teeth on the front and rear sprockets, respectively, Fig. 10–71, and Rբ and Rᵣ their respective radii. The teeth are spaced the same on both sprockets and the rear sprocket is firmly attached to the rear wheel.

Diagram showing the relationship between the angular velocities of a bicycle's rear wheel and front sprocket, with labeled components.

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. Consider a conveyor belt system that has two sprockets connected by a single belt. The driving sprocket which is attached to a motor has TD of teeth and a radius of RD. And the driven sprocket which is attached to the conveyor belt has TF of teeth and has a radius of RF. When the motor runs, the driving sprocket rotates with an angular velocity of WD. And the driven sprocket rotates with an angular velocity of WF, determine how the angular velocity of the driven sprocket. WF is related to the angular velocity of the driving sprocket WD by using the value for the number of teeth located on the sprockets. So that's our angle. Our angle is we're trying to figure out how the angular velocity of the driven sprocket is related to the angular velocity of the driving sprocket by using the value for the number of teeth located on the sprockets themselves. So that's our final answer. That we're ultimately trying to solve for. So with that in mind, let's look at the figure that's provided to us by the prom itself. In the far left at the bottom corner, we have our motor that's denoted by this black and blue rectangular shaped object which is our motor and the motor is connected to two sprockets. So the driving sprocket appears to be a larger gear, whereas the driven sprocket on the right is a smaller looking gear. And it's moving, this belt is moving in a counterclockwise motion because we have our vector V velocity pointing to the left. So it goes a left around our bigger sprocket around to our smaller sprocket then up and goes around in this counterclockwise motion. So we have our driving sprocket which has an angular velocity of WD. And then we have our angular velocity of our driven sprocket which is WF and then we have our radius of our driving sprocket which is RD and then our radius of our driven sprocket which is RF Awesome. So with that in mind, let's read off our multiple choice answers to see what our final answer might be. And let us note that for all of our multiple choice answers, it states that WF divided by WD is equal to sum expression written using units of teeth. So A would be two multiplied by TD, divided by TFB would be TD divided by TFC would be TD divided by two multiplied by TF and D would be TD divided by four multiplied by TF. OK. So first off, let us note that in order to solve this problem, we will need to consider the magnitudes of the linear velocity and the angular velocities of the sprockets. Now, at this point, we need to focus on the linear speed of the belt. Let us recall and note that the linear speed V of the belt is the same as it passes over both sprockets. Thus, we can recall and write the equation to sulfur V as V. The linear, the linear speed is equal to the angular velocity multiplied by R where R of course is the radius of the sprocket. And once again, W is the angular velocity. So now we need to equate the linear speeds. So let us recall and note that for the driving sprocket VD is so we need to recall and use the VC or use the equation for the driving sprocket velocity which states that VD is equal to WD multiplied by RD. And let us note for the driven sprocket. Similarly, the driven sprocket, we can write this equation as VF is equal to W F multiplied by RF Fantastic. So now since we know that the linear speed of the belt is the same, we can recall and write our angular velocity equation as WD multiplied by RD is equal to WF multiplied by RF. So now at this point, we need to recall and use the equation that considers the radius R of the sprocket that is related to the number of teeth capital T and the spacing distance D which is T multiplied by D. So T multiplied by lowercase D is equal to two multiplied by pi multiplied by capital R. So now we need to rearrange this equation to isolate and solve for R which will give us that R is equal to T multiplied by D divided by two multiplied by pi. Thus, we can write for the driving sprocket and the driven sprocket that R is so RD is equal to TD multiplied by D divided by two multiplied by Pi. And that RF is equal to TF multiplied by D divided by two multiplied by pi. OK. So now, at this point, we need to substitute the radius values into our angular velocity equation for both the driving sprocket and the driven sprocket. And when we do that, we will find that WD multiplied by TD multiplied by D divided by two multiplied by pi is equal to WF multiplied by TF multiplied by D divided by two multiplied by pi. So this will simplify to WD multiplied by TD is equal to WF multiplied by TF Fantastic. So we can simplify one last time to find our final answer. So when we do a little bit of rearranging here using some algebra, we will find that WF divided by WD is equal to TD divided by TF. And that's it. We saw for our final answer. Hooray, we did it. So with that in mind, looking at our multiple choice answers, the correct answer has to be the letter BWF, divided by WD is equal to TD, divided by TF. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Ah.

Key Concepts

Angular Velocity

Gear Ratio

Mechanical Advantage

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