Question

Ans: $\omega_{CD} = 0.5 \text{ rad/s}$ $\circlearrowright$ $\alpha_{CD} = 1 \text{ rad/}s^2 $\circlearrowright$$ Both disks A and B roll without slipping as shown. The telescope link CD connects the two rolling disks. Determine the angular velocity and the angular acceleration of the telescope link CD at the instant. $\omega_A = 1 \text{ rad/s}$ $R = 5'$ $\alpha_A = 3 \text{ rad/}s^2$ 10' $\omega_B = 2 \text{ rad/s}$ $\alpha_B = 3 \text{ rad/}s^2$ Rolling Rolling

          Ans: $\omega_{CD} = 0.5 \text{ rad/s}$ \(\circlearrowright\)
$\alpha_{CD} = 1 \text{ rad/}s^2 \(\circlearrowright\)$
Both disks A and B roll without slipping as shown. The telescope link CD
connects the two rolling disks. Determine the angular velocity and the angular
acceleration of the telescope link CD at the instant.
$\omega_A = 1 \text{ rad/s}$
$R = 5'$ 
$\alpha_A = 3 \text{ rad/}s^2$
10'
$\omega_B = 2 \text{ rad/s}$
$\alpha_B = 3 \text{ rad/}s^2$
Rolling
Rolling

Ans: ωCD = 0.5 rad/s ↻
αCD = 1 rad/s^2 ↻
Both disks A and B roll without slipping as shown. The telescope link CD
connects the two rolling disks. Determine the angular velocity and the angular
acceleration of the telescope link CD at the instant.
= 1 rad/s
R = 5'
= 3 rad/s^2
10'
= 2 rad/s
= 3 rad/s^2
Rolling
Rolling

Added by Ramon T.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Christopher Dzorkpata

Step 1

First, let's determine the angular velocity of disk A. We know that the angular velocity of a rolling disk without slipping is given by ω = v/r, where v is the linear velocity and r is the radius of the disk. Since the linear velocity of disk A is given as 3 m/s Show more…

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Ans: Ï‰ = 0.5r/s Both disks A and B roll without slipping as shown. The telescope link CD connects the two rolling disks. Determine the angular velocity and the angular acceleration of the telescope link CD at the instant. R = 5 A = 3 10 A