At the instant shown, the rod R is rotating about its center of rotation with =2
ID: 1791732 • Letter: A
Question
At the instant shown, the rod Ris rotating about its center of rotation with =2.5rad/s.
mA=10.6kg;
The pulley, with mP=10.7kg and RP=0.2m, may be modelled as a uniform disc.
The rod, with mR=6.2kg and L=0.8m, may be modelled as a thin beam rotating about one end.
g=9.8m/s ².
What is the magnitude of the acceleration of point B at this instant?
At the instant shown, the rod Ris rotating about its center of rotation with =2.5rad/s.
mA=10.6kg;
The pulley, with mP=10.7kg and RP=0.2m, may be modelled as a uniform disc.
The rod, with mR=6.2kg and L=0.8m, may be modelled as a thin beam rotating about one end.
g=9.8m/s ².
Explanation / Answer
Point B is moving in the circle about the fixed end of the rod.
It's instant acceleration towards the center, that is axis of rotation, = w2 L = 5 m/sec2
But besides this it also has acceleration in the tangent , that is in vertcal direction, at this instant of time. Let this acceleration be 'a' downward. Acceleration of mass A, by constraint is 'a' in the upward direction. Taking tension in string as T, for block A
T - Ma g = Ma (a)
T - 103.9 = 10.6 a ....1
For point B, a = alpha*L ......2 ( alpha = angular accelration of rod)
alpha = torque on rod / I ....3 ( I is moment of inertia of rod = Mr L2 /3 = 1.32 )
torque = Mr g L/2 - T L = 24.3 - 0.8 T .....4
combining 2 , 3 and 4 we get
1.65 a = 24.3 - 0.8 T ....5
eliminatin g T form 1 and 5 , we get
a = -5.8 m/s2 ( -ve sign indicates direction of acceleartion is opposite to the direction assumed, that is upward)
Total acceration of point B = root(a2 + ac2 ) (ac is acceleration toward center)
= 7.66 m/sec2
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