Question

the question is in last picture. i provided the lab content... I need guidance. thank you.

INVESTIGATION 10 ROTATIONAL MOTION OBJECTIVE To determine the moment of inertia I of a heavy composite disk by plotting measured values of torque versus angular acceleration. THEORY Newton's second law states that for translational motion (motion in a straight line) an unbalanced force on an object results in an acceleration which is proportional to the mass of the object. This means that the heavier the mass, the more it will resist changes in its velocity. We write the 2nd law as: Femam dv/dt. For rotational motion, there is an exact analog for each term in Newton's second law. The equation governing rotational motion la=1 do/dt (10.1) where I'net is the net torque on the body. Remember that torque is defined as I =rxF, 1 is the moment of inertia of the body, and a is the angular acceleration expressed in radians/s?. (Please Note: in some texts torque is indicated by t rather than r.) Let us think about I, the rotational inertia. I depends not only on the mass of the object on which a torque is acting, but also on the distribution of the mass about the axis of rotation. In fact / is defined by: I= m,n? (10.2) where m; is the mass of a particle, and is its distance from the axis of rotation. Equation (10-2) applies to system of particles. The contribution of each particle to the rotational inertia depends not only on the mass, but on the distance the mass is located from the rotational axis set-up similar to ours is shown in figure 10.1. However, in ours a heavy omposite disk is sitting horizontally on the table and is free to rotate about s central axis. A mass is suspended from a string which is wrapped round a central aluminum hub. If this mass is heavy enough to overcome ne friction of the disk's bearings, it will accelerate the disk/hub. We have a combination of motions in this problem. The mass 'm! Ecelerates downward with translational acceleration a. At the same time e tension T in the string applies a torque to the disk/hub assembly. Alding the forces acting on 'm' we have: Fy = mg - T = may (10.3) ence the tension in the string is given by: T = m (g - ay) Fig.10.1

Answer 1

2. If you apply a constant force (F) or a mass hanging from the string will have same effect as the acceleration due to gravity will not change for a short distance. Though you have considered constant g when you take mass m. Therefore, placing a constant mass is equivalent to constant force applied on it. Hence, the angular acceleration will not change. Therefore, F = T

3. The total torque is

$\alpha =\frac{a_{y}}{r_{0}}=\frac{1}{r_{0}}\left ( \frac{2y}{t^{2}} \right )$

$\tau _{T}=Tr_{0}=m\left ( g-a_{y} \right )r_{0}\ and\ a_{y}=\frac{2y}{t^{2}}$

$\tau _{T}=Tr_{0}=m\left ( g-\frac{2y}{t^{2}} \right )r_{0}$

For the heaviest masses, the mass will fall rapidly than the lighter masses. So, you have to measure, the time (t) and displacement (y) should be measured carefully, otherwise there will be much error in total torque (as it is multiplied by a factor m which is large) and for angular acceleration too. From the above equations support this comment.