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.

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written 8.4 years ago by

teamques10 ★ 68k

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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 =2 \\ BI = 2 IA\text {______}(i) \\ Also, BI + IA = BA(\text {diameter }) = 0.48\text {____}(ii) $

From (i) & (ii),

$BI = 0.32 m $ & $IA = 0.16 m $

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 = 9.375 rad/s \\ Vc = w * IC = 9.375*0.08 = 0.75 m/s $

Now, for calculation of VD , we need ID

Applying cosine rule,

$ID = \sqrt{[(0.24)2+(0.08)2 - 2*(0.24)(0.08)\cos135] }= 0.315 m $

Now, $VD = w * ID = 9.375 * 0.315 = 2.953 m/s $

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

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