Question

The figure shows a rod \(PQ\), hinged at \(P\), rotating counter-clockwise with angular speed \(\omega = 15 \, rad/s\). A block \(R\) translates along a slot cut in rod PQ. At the instant shown:
- Distance \(PR = 0.5 \, m\),
- \(\theta = 60^\circ\),
- Relative velocity of \(R\) with respect to PQ = 5 m/s,
- Relative acceleration of \(R\) with respect to PQ = 0.

Which one of the following is the CORRECT magnitude of the absolute acceleration (in m/s\(^2\)) of block R?

• A. 135.2
• B. 187.5
• C. 112.5
• D. 150.0

Hint:

For sliding along a rotating rod: - Coriolis acceleration = \(2 \omega v_{rel}\), - Centripetal acceleration = \(\omega^2 r\). They are always perpendicular, so take vector sum.

Answer 1

The Correct Option is B

Step 1: General acceleration relation in rotating frames.
For a particle sliding along a rotating rod: \[ \vec{a}_R = \vec{a}_{rel} + 2 \vec{\omega} \times \vec{v}_{rel} + \vec{\alpha} \times \vec{r} + \vec{\omega} \times (\vec{\omega} \times \vec{r}) \] Here, - \(\vec{a}_{rel} = 0\) (given), - \(\alpha = 0\) (uniform angular speed), - Only Coriolis and centripetal terms remain.

Step 2: Coriolis acceleration.
\[ a_{cor} = 2 \omega v_{rel} = 2 (15)(5) = 150 \, m/s^2 \]

Step 3: Centripetal acceleration.
\[ a_{c} = \omega^2 r = (15^2)(0.5) = 225 \times 0.5 = 112.5 \, m/s^2 \]

Step 4: Resultant acceleration.
These two are perpendicular (Coriolis tangential, centripetal radial). So: \[ a = \sqrt{a_{cor}^2 + a_{c}^2} = \sqrt{150^2 + 112.5^2} \] \[ a = \sqrt{22500 + 12656.25} = \sqrt{35156.25} \approx 187.5 \, m/s^2 \] Final Answer:
\[ \boxed{187.5 \, m/s^2} \]