A spring (k=250n/m) of natural length 0.10m is used to hold a 0.50kg mass into a

Question

A spring (k=250n/m) of natural length 0.10m is used to hold a 0.50kg mass into a circular motion with one end secured to rotation axis?

Ignore gravity and any other forces in the problem

1) What is the rotational period when the mass rotates at a radius of 0.2m?
2) What is the total mechanical energy of the system at that radius of rotation?
3)how much work would be required to increase the rotational radius from 0.2m to 0.3m?
4) In 3), is it more than, less than, or equal to the work needed to stretch the spring from 0.2m to 0.3m in length? Explain.

Explanation / Answer

1)

Fe = m*r*w^2

250*(0.2-0.1) = 0.5*0.2*w^2

w1 = 15.8 rad/s

time period T = 2pi/w1 = (2*pi)/15.8 = 0.398 s

2)

TE = 0.5*m*r*w^2 + 0.5*k*dx^2

TE1 = (0.5*0.5*0.2*15.8^2)+(0.5*250*0.1^2) = 13.732 J

3)

Fe = m*r*w^2

250*(0.2-0.1) = 0.5*0.3*w2^2

w2 = 13 rad/s

at r = 0.3 m

TE2 = (0.5*0.5*0.3*13^2)+(0.5*250*0.2^2) = 13.45 J

Work = 17.65 - 13.732 = 3.918

4)

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