Answer:
α = 7.848 rad/s^2 ... Without disk inertia
α = 6.278 rad/s^2 .... With disk inertia
Explanation:
Given:-
- The mass of the disk, M = 5 kg
- The right hanging mass, mb = 4 kg
- The left hanging mass, ma = 6 kg
- The radius of the disk, r = 0.25 m
Find:-
Determine the angular acceleration of the uniform disk without and with considering the inertia of disk
Solution:-
- Assuming the inertia of the disk is negligible. The two masses ( A & B ) are hung over the disk in a pulley system. The disk is supported by a fixed support with hinge at the center of the disk.
- We will make a Free body diagram for each end of the rope/string ties to the masses A and B.
- The tension in the left and right string is considered to be ( T ).
- Apply newton's second law of motion for mass A and mass B.
ma*g - T = ma*a
T - mb*g = mb*a
Where,
* The tangential linear acceleration ( a ) with which the system of two masses assumed to be particles move with combined constant acceleration.
- g: The gravitational acceleration constant = 9.81 m/s^2
- Sum the two equations for both masses A and B:
g* ( ma - mb ) = ( ma + mb )*a
a = g* ( ma - mb ) / ( ma + mb )
a = 9.81* ( 6 - 4 ) / ( 6 + 4 ) = 9.81 * ( 2 / 10 )
a = 1.962 m/s^2
- The rope/string moves with linear acceleration of ( a ) which rotates the disk counter-clockwise in the direction of massive object A.
- The linear acceleration always acts tangent to the disk at a distance radius ( r ).
- For no slip conditions, the linear acceleration can be equated to tangential acceleration ( at ). The correlation between linear-rotational kinematics is given below :
a = at = 1.962 m/s^2
at = r*α
Where,
α: The angular acceleration of the object ( disk )
α = at / r
α = 1.962 / 0.25
α = 7.848 rad/s^2
- Take moments about the pivot O of the disk. Apply rotational dynamics conditions:
Sum of moments ∑M = Iα
( Ta - Tb )*r = Iα
- The moment about the pivots are due to masses A and B.
Ta: The force in string due to mass A
Tb: The force in string due to mass B
I: The moment of inertia of disk = 0.5*M*r^2
( ma*a - mb*a )*r = 0.5*M*r^2*α
α = ( ma*a - mb*a ) / ( 0.5*M*r )
α = ( 6*1.962 - 4*1.962 ) / ( 0.5*5*0.25 )
α = ( 3.924 ) / ( 0.625 )
α = 6.278 rad/s^2
