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Due to slipping, points A & B on the rim of disk have the velocities as shown in fig. Determine the velocities of the centre point C & point D on rim at this instant.

Take radius of disk 0.24m

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enter image description here

Consider above diagram in which BX & AY are velocity vectors

AB is diameter of circle & C is ICR of disk.

Now, triangles XBI & IAY are similar

Therefore, BI/IA=BX/AY=3/1.5=2BI=2IA______(i)Also,BI+IA=BA(diameter )=0.48____(ii)

From (i) & (ii),

BI=0.32m & IA=0.16m

Now, as BI=0.32 & BC=0.24,

IC(distance between centre of circle & ICR) =0.08

Now, w of disc is VB/IB=3/0.32=9.375rad/sVc=wIC=9.3750.08=0.75m/s

Now, for calculation of VD , we need ID

Applying cosine rule,

ID=[(0.24)2+(0.08)22(0.24)(0.08)cos135]=0.315m

Now, VD=wID=9.3750.315=2.953m/s

Velocity of C is 0.75m/s & velocity of D is 2.953m/s

enter image description here

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