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GATE 2024 ME – Question 24

Engineering Mechanics · Impulse and momentum, energy formulations · 1 mark · Multiple choice

A ram in the form of a rectangular body of size $l = 9$ m and $b = 2$ m is suspended by two parallel ropes of lengths 7 m. Assume the center-of-mass of the body is at its geometric center and $g = 9.81$ m/s$^2$. For striking the object P with a horizontal velocity of 5 m/s, what is the angle $\theta$ with the vertical from which the ram should be released from rest?

A rectangular ram hanging from two parallel ropes of length 7 m, each making an angle $\theta$ with the vertical. The ram is shown at its lowest position (dashed) next to the object P.
  1. 67.1°
  2. 40.2°
  3. 35.1°
  4. 79.5°

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Correct answer: (C) 35.1°

Explanation

Two equal parallel ropes make the ram move without turning, so every point of it moves at the same speed and it behaves as a pendulum bob of length 7 m. It falls through a height $h = 7(1 - \cos\theta)$, and at the lowest point $\frac{1}{2}v^2 = gh$. So $25 = 2 \times 9.81 \times 7(1 - \cos\theta)$, which gives $1 - \cos\theta = 0.1820$, $\cos\theta = 0.8180$ and $\theta = 35.1^\circ$.