Question

>15N ТА A force of 150 is applied on a disk as shown in figure. The moment of inertia of disk about its center of mass is į mot. The point of rotation and center of mass are the same that is point G in this problem If the radius of disk is 0.5m and the mass is 2kg find the d that is angular accelaration of the disk: If disk starts from rest, find wo that is angular speed after 3 seconds. Do this question by impulse momentum relation. And then do it again by using Newton's method. Also find the and accelaration point A. Find the and accelaration of point G. There is only one pin joint in this figure that is point G. The point A is just made because of the question that is asked in the Problem 1. Point A is not a pin joint, is not a joint, is nothing. Point A is at -0.5 i (i is unit vector) with respect to point velocity the velocity G.

Answer 1

Given : F-15N m = 2 kg at 0.5 m R IG= mo? = 4* 2x(0.5) G let Al An 0.25 kg-m2 be the angular acceleration Torque I x x FXR = I x x = 15x0.5 = 0.25XX x = 30 rad/s² Now Calculating angular velocity of dise after 3 sec Using Impulse - momenterm relation Torque x at I (W-Wol; here Wo=0 15X0.5x3 0.25(6-0) ► W = 90 ralls Using Newton's method : W=W.+2xt w = 04 30x3 = 90 rad/s Now calculating Velocity and acceleration of point A. Initially at t=0 VA = 0 Now acceleration : Normal acceleration at point A VA² an R R Tangential acceleration at = Rxa=0.5830= 15 rad/s2 2.

As point G is fixed thus acceleration and velocity point G will be of Zero VG O and AG
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