Question Video: Calculating the Centripetal Force of a Unit Mass That Traces a Circular Path of a Unit Diameter in a Unit Second | Nagwa Question Video: Calculating the Centripetal Force of a Unit Mass That Traces a Circular Path of a Unit Diameter in a Unit Second | Nagwa

Question Video: Calculating the Centripetal Force of a Unit Mass That Traces a Circular Path of a Unit Diameter in a Unit Second Physics • First Year of Secondary School

What is the magnitude of the centripetal force that must act on an object of mass 1.0 kg to make it move along a circular path of diameter 1.0 m, completing a circle every 1.0 s?

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Video Transcript

What is the magnitude of the centripetal force that must act on an object of mass 1.0 kilograms to make it move along a circular path of diameter 1.0 meters, completing a circle every 1.0 seconds?

Well, okay, so in this scenario, we have a circular path. And we’re told that the diameter of this path is 1.0 meters. Along with this, we have an object whose mass, that we can call 𝑚, is given as 1.0 kilograms. We want this object to move around the circular path so that it completes a circle every 1.0 seconds. One time around the circle is a complete revolution. So that means we can refer to this time of 1.0 seconds as our period of revolution. The question is, in all this, what’s the magnitude of the centripetal force that must act on this mass? That is, for the mass to be moving in a circle, there must be some force making it tend towards the center of that circle. That’s the centripetal force, and we want to figure out how strong it is.

We can start doing this by recalling that the centripetal force on an object is equal to the object’s mass times its centripetal acceleration. Now, an object’s centripetal acceleration is given by its speed squared divided by the radius of the circular path it moves in. But then we can also recall that for an object moving in a circle, its linear speed, its tangential speed around that circle, is equal to the circle’s radius times the object’s angular speed 𝜔. If we substitute 𝑟 times 𝜔 in for 𝑣 in our equation for centripetal acceleration, we find it’s equal to the quantity 𝑟 times 𝜔 squared divided by 𝑟 or, more simply, 𝑟 times 𝜔 squared. We can then substitute 𝑟 times 𝜔 squared in for 𝑎 sub c in our equation for centripetal force. And now we have an expression for this force that we can work with in terms of our given parameters.

Starting with our object’s mass, we’re given that it’s 1.0 kilograms. And then the radius of the circle our object is moving in is equal to the diameter divided by two. And then lastly, there’s this factor here, the angular speed 𝜔 squared. We can recall that the units of 𝜔 are radians per second. And since we’re told that our object completes a circle every 1.0 seconds, recalling that a full circle is equal to two 𝜋 radians, we can say that the angular speed of our object is equal to two 𝜋 radians divided by 1.0 seconds. This is its angular speed. So then, substituting in to our equation for centripetal force, here’s what we get. Our mass as we saw is 1.0 kilograms. Our circle’s radius is its diameter, 1.0 meters, divided by two. That’s 0.5 meters. And then our object’s angular speed is two 𝜋 radians in 1.0 seconds. And this factor of 𝜔 is squared.

When we calculate all this out, we find a result of 19.73 and so on and so forth newtons. But notice how all of the values we’re given in our problem statement have two significant figures. That’s how many we’ll keep in our answer. And to two significant figures, our result is equal to 20 newtons. That’s the magnitude of the centripetal force acting on our object.

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