Question Video: Center of Mass of a System of Objects Undergoing Circular Motion | Nagwa Question Video: Center of Mass of a System of Objects Undergoing Circular Motion | Nagwa

# Question Video: Center of Mass of a System of Objects Undergoing Circular Motion

Two particles of masses 2.0 kg and 4.0 kg move in uniform circles with radii of 5.0 cm and π cm respectively. The π₯-coordinate of the particle moving in the 5.0 cm radius circle is given by π₯(π‘) = 5.0 cos (2π‘) and the π¦-coordinate is given by π¦(π‘) = 5.0 sin (2π‘). The π₯-coordinate of the center of mass of the particles is given by π₯_(cm) (π‘) = 6.0 cos (2π‘) and the π¦-coordinate of the center of mass of the particles is given by π¦_(cm) (π‘) = 6.0 sin (2π‘). Find π.

04:30

### Video Transcript

Two particles of masses 2.0 kilograms and 4.0 kilograms move in uniform circles with radii of 5.0 centimeters and π centimeters, respectively. The π₯-coordinate of the particle moving in the 5.0-centimeter radius circle is given by π₯ as a function of π‘ equals 5.0 cos of two π‘. And the π¦-coordinate is given by π¦ as a function of π‘ equals 5.0 sin of two π‘. The π₯-coordinate of the center of mass of the particles is given by π₯ sub cm as a function of π‘ equals 6.0 cos of two π‘. And the π¦-coordinate of the center of mass of the particles is given by π¦ sub cm as a function of π‘ equals 6.0 sin of two π‘. Find π.

We want to solve for the radius of rotation of our 4.0-kilogram particle. And to do that, we can start by drawing a diagram of this scenario. In this example, weβre told about two masses that are moving in circles. Mass one moves in a circle of radius weβve called π of 5.0 centimeters. The second mass, mass two, moves in a circle of radius capital π that we want to solve for.

With respect to the smaller mass, weβre told the π₯- and π¦-coordinates of that mass as a function of time. And weβre also told the π₯- and π¦-coordinates of the center of mass of this system of two masses as a function of time. Even with the center of mass that changes with time, the relationship that the center of mass is equal to the sum of the product of each mass element multiplied by its distance from the center of mass divided by the sum of the overall masses still applies to our situation.

Letβs look for a moment at the π₯-position of our particle and the π₯-position of the center of mass of the system. We notice that these two expressions have the same phase. And if we look at π¦ as a function of π‘ and the center of mass π¦-coordinate, we see the same relationship that the phase is the same. This means we can solve for the radius of our second particle, capital π, by looking at either the π₯-coordinate or separately the π¦-coordinate. Both will give us the same information.

Just to choose one of the coordinates, letβs write down the center of mass relationship in the π₯-direction. The π₯-coordinate of our system of masses center of mass is equal to π one times π₯ as a function of π‘ times π two times the function weβve called π₯ sub two as a function of π‘, currently unknown, all divided by the sum of π one and π two. Letβs plug in to this equation what we know. We know π₯ sub cm, π one, π two, and π₯ as a function of time.

When we write in all the information we know, we see that the final form of our center of mass has an expression cos of two π‘. We see that same form in the expression of the exposition of our smaller particle. And knowing that the 5.0 that precedes that phase information represents the radius of our smaller particleβs rotation. This gives us a clue that the expression for π₯ sub two as a function of π‘ will be the radius of the larger particle, capital π, multiplied by the same phase, cos of two π‘.

We know that π₯ sub two of π‘ will have this form because this is the exposition of the particle with a radius capital π. And we know the phase relationship of this second particleβs position must be consistent with the phase relationship of the overall center of mass of the system. So substituting this expression into our center of mass equation, we see that the units of kilograms cancel out from this expression. And if we multiply both sides of our equation by 6.0 and then subtract 10.0 times the cos of two π‘ from both sides and then seeing the cos of two π‘ appearing on both sides canceling that trigonometric phase term out. We see that 26 is equal to 4.0π or π is equal to 26 divided by 4.0, where π implicitly is in units of centimeters. To two significant figures as a decimal, π is equal to 6.5 centimeters. Thatβs the radius of the circle in which the larger particle moves.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions