Video: Calculating the Speed of an Object Undergoing Circular Motion

A 60 kg rock is attached to the end of a piece of string and is rotated in a circle. If the radius of this circle is 10 m and the rock makes one complete revolution every 5.0 s, then which of the following is nearest to the speed of the rock? [A] 13 m/s [B] 12 m/s [C] 4 m/s [D] 15 m/s [E] 6 m/s

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Video Transcript

A 60-kilogram rock is attached to the end of a piece of string and is rotated in a circle. If the radius of this circle is 10 meters and the rock makes one complete revolution every 5.0 seconds, then which of the following is nearest to the speed of the rock? (a) 13 meters per second, (b) 12 meters per second, (c) four meters per second, (d) 15 meters per second, (e) six meters per second.

The question asks us for the speed of a rock as it moves in a circle, given the mass of the rock, the radius of the circle, and the time it takes to make one complete revolution. Let’s draw a diagram to help organize our thoughts. In this diagram, we’ve included the piece of string which forms the radius of the circle, the rock, and the circular path that is the motion of the rock. We’ve also labeled the diagram with symbols and values from the question.

The radius of the circle with the symbol π‘Ÿ is 10 meters per second. The mass of the rock with the symbol π‘š is 60 kilograms. The time for one revolution with the symbol capital 𝑇 is 5.0 seconds. And the speed of the rock with the symbol 𝑣 is what we’re looking for. Since this question is looking for the speed of the rock, let’s start by recalling the definition of speed. The speed of an object, 𝑣, is equal to the distance traveled by the object, Δ𝑑, divided by the time it took to travel that distance, Δ𝑑.

Looking at our diagram, we can see that we already have a length of time. Capital 𝑇, 5.0 seconds, is the time it takes for the rock to travel once around the circle. If we take capital 𝑇 to be our interval of time, then we need the distance that the rock travels in this time. But that distance is just one complete revolution around the circular path, in other words, the circumference, two πœ‹ times the radius. Now, we have a distance and the corresponding length of time. So we’ll take Δ𝑑 divided by Δ𝑑. And that’s the speed of the rock. So we have that two πœ‹π‘Ÿ divided by capital 𝑇 is equal to Δ𝑑 divided by Δ𝑑, the speed. Let’s now plug in values from the question.

Additionally, we’ll also use the approximate value 3.14 for the constant πœ‹. So we have two times 3.14 times 10 meters divided by 5.0 seconds. Two times 3.14 is 6.28. 10 meters divided by 5.0 seconds is two meters per second. And 6.28 times two meters per second is 12.56 meters per second. Remember, though, that the question wanted the value nearest to the speed of the rock, and all of the choices are integers. So we must round 12.56 to the nearest integer. 12.56 is closer to 13 than it is to 12. So 12.56 meters per second is approximately 13 meters per second. We therefore see that the answer choice nearest to the speed of the rock is (a), 13 meters per second.

It’s worth briefly mentioning another way we could’ve approached this problem that works specifically because the motion is in a circle. For an object moving in a circle, we also have that 𝑣, the speed, is equal to πœ”, a quantity called the angular frequency, times π‘Ÿ, the radius. πœ” is equal to two πœ‹ divided by capital 𝑇, the period of the motion. To understand why πœ” is called the angular frequency, recall that there are two πœ‹ radians in a complete circle. So the quantity two πœ‹ divided by capital 𝑇 is the total angles in one cycle divided by the total time for that cycle, and so the name angular frequency.

At any rate, if we plug in πœ” is equal to two πœ‹ divided by capital 𝑇 into 𝑣 equals πœ”π‘Ÿ, we get 𝑣 is equal to two πœ‹ divided by capital 𝑇 times π‘Ÿ. But this result for speed is equivalent to our result from before, two πœ‹π‘Ÿ divided by capital 𝑇. Therefore, when we plug in values for π‘Ÿπœ‹ and capital 𝑇, we should get the same result. And again, we find that the speed of the rock is approximately 13 meters per second. Thus, whether we use 𝑣 equals πœ”π‘Ÿ, which is unique to rotational motion, or 𝑣 equals Δ𝑑 divided by Δ𝑑, the general definition of speed, we still get the same result.

As one final note, neither of these two approaches relied on the fact that the mass of the rock is 60 kilograms. And this is because the speed of an object doesn’t depend on its mass. It only depends on the distance the object travels and the time it takes to travel that distance.

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