Question

A meterstick (L = 1 m) has a mass of m = 0.134 kg. Initially, it hangs from two short strings: one at the 25 cm mark and one at the 75 cm mark.

1)

What is the tension in the left string? N

2)

Now the right string is cut! What is the initial angular acceleration of the meterstick about its pivot point? (You may assume the rod pivots about the left string, and the string remains vertical) rad/s^2

3)

What is the tension in the left string right after the right string is cut? N

4)

After the right string is cut, the meterstick swings down to where it is vertical for an instant before it swings back up in the other direction.

What is the angular speed when the meterstick is vertical? rad/s

5)

What is the acceleration of the center of mass of the meterstick when it is vertical? m/s^2

6)

What is the tension in the string when the meterstick is vertical? N

I need help with 4-6, thank you!!

Answer 1

mass of the stick, m=0.134 kg

L=1m

x1=25cm , x2=75 cm

1)

Tension , T=m*g/2

=0.134*9.8/2

=0.657 N

2)

Net torque=I*alpa

m*g*x1=(m*L^2/12 + m*x1^2)*alpa

g*x1=(L^2/12 + x1^2)*alpa

9.8*0.25=(1^2/12 + 0.25^2)*alpa

===> alpa=16.8 rad/sec^2

3)

here,

T=m(g+a)

T=m*(g+x1*alpa)

=0.134*(9.8+0.25+16.8)

=3.6 N

4)

by using energy relation,

1/2*I*w^2=m*g*x1

1/2*(m*L^2/12 + m*x1^2)*w^2=m*g*x1

1/2*(0.134*1^2/12 + 0.134*0.25^2)*w^2=0.134*9.8*0.25

===> w=5.8 rad/sec

5)

a_cm =w^2*x1

=5.8^2*0.25

=8.41 m/sec^2

6)

T=m*(g+a_cm)

=0.134*(9.8+8.41)

=2.44 N