A 0.8 kg mass is moving in a circle of radius 1.6 m on a flat frictionless table at the end of a string. The speed of the mass is 3.6 m/s. The string routes through a hole in the center of the table and is held by you underneath the table.
What is the angular momentum of the mass?
Express your answer using three significant figures.
I found this to be 4.61 kg m^/s
I just cant figure out the rest of the problems! Someone please help!
What is the tension in the string?
Express your answer using three significant figures.
Tension = _________N
** I tried using T-mg= ma and finding a using the equation a= v^2/r but it was wrong. What am I doing wrong?!
If you pull on the string so that the radius of the circle decreases to one half its former value, what is the new tension in the string?
Express your answer using three significant figures.
Tension = _________N
How much work did you do to reduce the radius by a factor of one-half?
Work = ________J
Answers & Comments
Verified answer
L = m v r
= 0.8 kg*3.6 m/s * 1.6 m = 4.61 kgm^2/s
The centripetal force for the circular motion is delivered by the tension in the string:
T = m v^2/r
= 0.8 kg * (3.6 m/s)^2 / 1.6m = 6.48N
( you used T-mg, but the mass is on a table so the gravity is balanced by the normal force. Besides, gravity is not in the direction of motion in this case...)
because you pull centrally, you do not apply a torque. Therefore angular momentum is conserved. If you half the radius, you see from L= mvr that the velocity must double to keep L constant. But doubling velocity and halving r gives a factor 2^2/(1/2) = 8 in the tension.
Hence T = 8*6.48N=51.8N
The initial kinetic energy is 1/2 mv^2 . The final kinetic energy is 1/2 m (2v)^2.
So the work ( which here equals the increase in kinetic energy) is
W = 2 m v^2 - 1/2 m v^2 = 3/2 mv^2
= 3/2* 0.8 kg * (3.6m/s)^2
= 15.6 J
angular momentum of the mass is 4.61 kg m² /s
You apply a force and it is the tension in the string which provides the centripetal force
Tension = mv²/r = 0.8*3.6²/1.6 =6.48 N.
( why do you use "mg " here ?)
New tension in the string = mv²/ (r/2) = 2*6.48 = 12.96 N
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The excess force applied to move the mass is 6.48 N and it is moved through a distance of 0.8 m
work done is 6.48*0.8 = 5.184 J
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