Question

A meterstick (L = 1 m) has a mass of m = 0.168 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?

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)

3) What is the tension in the left string right after the right string is cut?

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?

5) What is the acceleration of the center of mass of the meterstick when it is vertical?

6) What is the tension in the string when the meterstick is vertical?

7) Where is the angular acceleration of the meterstick a maximum? Options are as follows:

a) right after the string is cut and the meterstick is still horizontal

b) when the meterstick is vertical at the bottom of its path

c)the angular acceleration is constant

PLEASE SHOW WORK THIS IS MY LAST TRY THANK YOU IN ADVANCE WILL GIVE A THUMBS UP IF RIGHT

Answer 1

Given that

Length of meterstick L = 1 m

Mass of stick m = 0.168 kg

RL (distance of left side part of masterstick) = 25 cm =0.25m

Rr (distance of right side part of masterstick)= 75 cm = 0.75 m

Part(a)

T = mg / 2 = 0.168*9.8 / 2 = 0.8232 N

Part(b)

Net torque = RL*F = RL*mg = 0.25*0.168*9.8 = 0.4116 N-m

We know that

Net torque = I .α

where α = angular acceleration of the meterstick

and inertia of masterstik I = m*L^2 / 12 + mRL^2

I = 0.168*(1)^2 / 12 +0.168*0.25^2

= 0.0245 kg.m2

α = /I

α = 0.4114 / 0.0245

α  = 16.79 rad / s2

Part(c)

when we remove right string net force acting on masterstick

Fd - T = F

mg - T = ma

here a = acceleration = α*RL = 16.79*0.25

a = 4.2 rad/s^2

0.168*9.8 - T = 0.168*4.2

T = 0.94 N

Part(d)

when the meterstick is vertical Work done by gravity on masterstick = change in kinetic energy

= Fd.RL= (1/2)* I*ω^2

= mg.RL = (1/2)* I*ω^2

0.168*9.8*0.25 = 0.5*0.0245*ω^2

ω = 5.8 rad / s

Part(e)

acceleration of the center of mass of the meterstick when it is vertical

Ac = RL*ω^2

Ac = 0.25*5.8^2

Ac= 8.41 m / s2

Good Lucck !!!