A ball travels on a frictionless circular track at 3m/s. The ball cannot have an acceleration greater than 1.5m/s 2 or it will go off the track. What is the smallest radius the circular track can have so that the ball stays on the track?

Guys. So let's get some practice now with uniform circular motion. We have this ball that's traveling on a frictionless circular track and we have the speed, we know this is just gonna be 3 meters per second. So remember, this is a tangential velocity. This is v t. It's 3. Now we're told that the ball cannot have an acceleration greater than 1.5 meters per second squared or it's gonna fall off the track. So we know that this ball, as it's going around its circular path, has a centripetal acceleration that points towards the center. And so basically, we wanna figure out what is the smallest radius. Right? What is the smallest radius of the circular track such that that acceleration isn't greater than 1.5 and the ball is gonna fall off the track. So, really, what we're trying to figure out here is what is r. So remember that there's a relationship between all three of these variables here, and it's basically just our AC equation. We know that a centripetal equals vtangential squared divided by r. So what we wanna do is if we're trying to figure out the minimum distance or the minimum radius, the smallest radius, that's gonna be related to the maximum centripetal acceleration. We know that acmax is gonna equal 1.5, so this is gonna equal vt squared divided by rmin, and that's just rmin. Right? Okay. So we wanna solve for this variable. All we have to do is just get it by itself. So we're basically just gonna flip the r and the a c. So now our r min becomes v tangential squared divided by a cmax. If the radius were any smaller than this minimum radius that we're calculating, then the acceleration is gonna be too high and the ball is gonna fly off the track. So basically, we just plug our numbers in. This is gonna be 3 squared divided by the, asymmetrical which is 1.5 and you're gonna get a radius of 6 meters. Alright. So anything smaller than that and the ball flies off the track. That's it for this one, guys.

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8. Centripetal Forces & Gravitation

Uniform Circular Motion

Physics

8. Centripetal Forces & Gravitation

Uniform Circular Motion

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Multiple Choice

A ball travels on a frictionless circular track at 3m/s. The ball cannot have an acceleration greater than 1.5m/s² or it will go off the track. What is the smallest radius the circular track can have so that the ball stays on the track?

R = 3 m

R = 12 m

R = 2 m

R = 6m

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Verified step by step guidance

Identify the type of motion: The ball is moving in a circular path, which means it is undergoing circular motion. In circular motion, the acceleration is centripetal and directed towards the center of the circle.

Use the formula for centripetal acceleration: The centripetal acceleration \( a_c \) is given by the formula \( a_c = \frac{v^2}{r} \), where \( v \) is the velocity of the ball and \( r \) is the radius of the circular path.

Substitute the given values: We know the velocity \( v = 3 \text{ m/s} \) and the maximum allowable centripetal acceleration \( a_c = 1.5 \text{ m/s}^2 \). Substitute these values into the formula to find the radius \( r \).

Rearrange the formula to solve for \( r \): \( r = \frac{v^2}{a_c} \). Substitute the known values into this equation to calculate the smallest radius.

Calculate the smallest radius: Perform the calculation \( r = \frac{3^2}{1.5} \) to find the smallest radius that will keep the ball on the track without exceeding the maximum acceleration.

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