If you experienced these lab questions, please help me out answer them. I apprec

Question

If you experienced these lab questions, please help me out answer them. I appreciate your help.


1. Use the definition of moment of inertia to derive the theoretical expressions for the moments of inertia of the disk and the ring.

2. Use Newton’s Second Law to derive the expression for the experimentally determined moment of inertia.

3. Use Work and Mechanical Energy to derive the expression for the experimentally
determined moment of inertia.

4. The average error in starting and stopping the stopwatch is about ±0.1 s. How much error would this contribute to your values for I ? What other sources of error can you think of? Remember that "human error" is not an acceptable excuse!

Explanation / Answer

The theoretical values of moment of inertia of the solid disc is

Id = (1/2) m R^2


m is the mass , R is the radius of the disc.

The moment of inertia of ring of outer radius Rb and inner radius Ra is

Ir = (1/2) m ( Rb^2 + Ra^2)


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2)The torque acting on the body which is moving with an angular acceleration a (alpha) when a force F is applied is


t ( tau ) = I a

it can also be defined as t (tau) = R x F ( here R is the radius , F is the force applied along the line tangential to the radius vector)

equating them I a = R F
moment of inertia is
I = RF / a
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3) The third procedure is bit different. we would take a object whose moment of inertia to be calculated at one end of a pulley and the other end is attached to a mass hanger. As the mass hanger moves downward, the gravitational potential energy is equal to the linear kinetic and rotational kinetic energies.

m g h = (1/2)m Vf^2 + (1/2) I ?^2 here ? (omega) is angular speed


The mass falls with uniform velocity, the velocity of the fall can be found with the height h
h = (Vf - Vi / 2 ) t .
Initial speed Vi is zero, so the above formula for the final speed is


Vf= 2 h / t
The angular velocity of the support is


?(omega) = Vf / R = 2h / Rt

now substituting in our main equation for moment of inertia

m g h = (1/2) m (2h/t)^2 + (1/2) I [ 2h/ Rt ]^2


I = mR^2 { [ gt^2 / 2h ] -1 }

------------------------------------------------------------- 4) Error in time is       t = 1 + 0.1 =1.10 s          = 1-0.10 s =0.9s The error in momenta of inertia is            I = mR^2 { [ gt^2 / 2h ] -1 }
Substitte t= 1.1 s and 0.9s we get the error in I.

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