Question

M, a solid cylinder (M=1.39 kg, R=0.111 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.570 kg mass, i.e., F = 5.592 N. Calculate the angular acceleration of the cylinder. (the answer to this is 39.8 rad/s^2) The cylinder is changed to one with the same mass and radius, but a different moment of inertia. Starting from rest, the mass now moves a distance 0.463 m in a time of 0.470 s. Find Icm of the new cylinder. I do not know how to answer this second part. Please help me if you can. Thank you.

M, a solid cylinder (M=1.39 kg, R=U.111 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.570 kg mass, i.e., F = 5.592 N. Calculate the angular acceleration of the cylinder. 7.25x101 rad/s^2 You are correct. Your receipt no. is 149-164 Previous Tries

Answer 1

Moment of inertia is the property of the body due to which it resists angular acceleration (same function is done by mass for linear motion), which is the sum of the products of the mass of each particle in the body with the square of its distance from the axis of rotation. Formula for Moment of Inertia can be expressed as:

I = Σ mari

A solid cylinder rotating on an axis that goes through the center of the cylinder, with mass M and radius R, has a moment of inertia determined by the formula:

$I=\frac{1}{2}MR^{2}$

In this problem, for a solid cylinder, M = 1.39 Kg and R = 0.111 m,

I =-MRP = -1.39 Kg x (0.111 m)2 = 0.008563095 Kg.m?

Torque or moment of a force is the tendency of a force to rotate the body to which it is applied and it is expressed by the multiplication of the magnitude of force and shortest distance between the rotational axis and the direction of the force. In this problem, the force F is applied to a cylinder of radius R. Hence, the torque ($\tau$) applied to the cylinder rotating along the axis goes through the centre

T= FR = 5.592 N X 0.111 m = 0.620712 N - m

Let us consider, the angular acceleration of the cylinder is $\alpha$, hence the torque according to Newton´s second laws of motion will be

$0.008563095\times \alpha =0.620712\\ \therefore \alpha=\frac{0.620712}{0.008563095}=72.487\ NKg^{-1}m^{-1}$

The angular acceleration of the cylinder is 72.487 NKg^-1m^-1.

Starting from rest, the mass now moves a distance 0.463 m in a time of 0.470 s. The mass moves in free-fall motion. From the equation of motion, , we can obtain the linear acceleration a,

H = fat?

2H + 2 x 0.463 0.3 m/s2 = 4.192 m/s 0.4702

You can calculate the I_CM from the idea given below:

The mass of the object is m = 0.570 Kg. The acceleration of the mass calculated above is a = 4.192 m/s² which is less than the acceleration due to gravity. Hence, the net acceleration is a = 4.192 m/s². Therefore, the net force applied on the cylinder in the vertical direction is F = m.a = 0.570 Kg x 4.192 m/s² = 2.38944 Kg.m.s²

Hence, the torque applied to the cylinder is T = F.R = 2.38944 Kg.m.s^-2 x 0.111 m = 0.265228 Kg.m²s^-2.

The moment of inertia of the cylinder is I_CM

The torque of the cylinder will be T = I_CM x $\alpha$ = I_CM a/R

Therefore,

$I_{CM}=\frac{T R}{a}=\frac{\left (0.265228\ Kg.m^{2}s^{-2} \right )\times 0.111\ m}{4.192\ m/s^{2}}=70.2297\times 10^{-4}\ Kg.m^{2}$