A 10gram point mass is attached to a string, and is traaveling in a circle at 2R
ID: 1684662 • Letter: A
Question
A 10gram point mass is attached to a string, and is traaveling in a circle at 2RPS, at a distance of 10cm from the center of rotation.1) What is the centripetal acceleration?
2) What is the cnetrifugal force?
3) What is the tention in the string?
4) What is the tangential velocity of the mass?
5) What is the KE of the mass ( using linear motion )
6) What is the KE of the mass ( using roatinal motion )
7) If the string ( i.e. the radius ) is immediately shortened to 5cm, and no KE is lost, What is the new angular speed ( given in RPS )
8) Now consider the rotating mass to be a sphere ( I=2/5 MR^2, with radius 0.5cm. The mass is still 10grams, and it is still traveling in a circle with a radius 10cm, ignore the string. Additionally, the small mass is rotating around its center at 10 Rps .
What is the total KEe system now
Explanation / Answer
2 revolutions per second means 4pi radians per second. Hence, angular speed W is 4pi. 1) centripetal acceleration is given by formula rW^2 where r is radius. Thus, answer is 10cm x 4pi = 0.1 x (4pi)^2 = 15.78. 2) centrifugal force is mass x centrifugal acceleration = 15.78 x 0.01Kg = 0.1578N. 3) It is the tension that provides the centripetal force in the motion. Thus, tension = 0.1578N. 4) tangential velocity is radius x centrifugal velocity = 0.1m x 4pi = 1.256m/s. 5)linear KE = 0.5mv^2 = 0.5 x 0.01 x 1.256^2 = 0.00788768 J. 6) rotational KE is 0.5IW^2 where I is moment of intertia which is mr^2 for point mass. Thus, energy = 0.5 x 0.01 x 0.1^2 x 16pi^2 = 0.00788768J. 7) Energy before and after shortening is same. Therefore, 0.5m(0.1^2)(16pi^2) = 0.5m(0.05^2)(W^2) where w is the new angular speed. Thus, W = 4 x 16pi^2 = 64pi^2 = 631.0144rad/s. 8) Now, angular speed is 20pi. Thus, translational velocity is 0.1 x 20pi = 2pi. Thus, translational KE is 0.5(0.01)(2pi)^2. Angular KE will be 0.5IW^2 = 0.5(0.4x0.01x0.05^2)(20pi)^2. Thus, total KE is 0.5(0.01)(2pi)^2 + 0.5(0.4x0.01x0.05^2)(20pi)^2. I can imagine only 1 place where you may have been confused. 1) In the formula translational velocity/acceleration = radius x angular velocity/acceleration, we use the radius of the CIRCULAR PATH (in this case the length of the string). NOT THE RADIUS OF THE SPHERE OR THE OBJECT.
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