Problem 2: Impulse-Momentum Consider a uniform slender rod with length L of 800

Question

Problem 2: Impulse-Momentum Consider a uniform slender rod with length L of 800 mm and mass m of S kg. The rod is suspended initially at rest from a hinge C. The rod is subject to an impulse of 6 Ns applied at 20P from the horizontal at D. Use r 100 mm and I-200 mm. The mass moment of inertia for a rod is Io ml2/12. L2 6 Ns 2.1 The mass moment of inertia about point Cis kgm2 (D)I 0.1 0510E) 10.285 1 ?10442] rad/s 2.2 The angular velocity immediately after the impulse is applied is (D) 6.98 (E) 5.34 (F) 2.41 2.3 The vertical component of the impulse at the pinC is- Ns A)10.837 (B) 1.16 | (C) | 0.232 D 2.43 (E) 0.514 (F) 2.4 After the impact, the rod will instantaneously gain velocity and will start rotating up until it rcaches the maximum heigh. The angle (measured from vertical) at this maximum height is (A) 85.5 B 89.1 (C) 245 (D) 1 62.8) (E) I 16. l | (F) 1 46.6

Explanation / Answer

2.1

Moment of inertia about G is mL2 / 12 = 5*0.82 / 12 = 0.2667 kg-m2

By parallel axis theorem, MOI about C is = 0.2667 + mr2 = 0.2667 + 5 * 0.12 = 0.317 kg-m2

2.2

Angular momentum = (6cos20) * (r+l) = (6cos20) * (0.1 + 0.2) = 1.69 kg-m2/s

Angular momentum = I*w

Angular velocity w = 1.69 / 0.317 = 5.34 rad/s

2.3

Vertical component of impulse = 6sin20 = 2.05 Ns

2.4

Kinetic Energy just after the impulse = 1/2 Iw2 = 1/2 * 0.317 * 5.342 = 4.517 J

Potential energy at the maximum height = mgr - mgr Cos theta = mgr (1 - cos theta) = 5 * 9.81 * 0.1 * (1 - cos theta)

By energy conservation,

4.517 = 5 * 9.81 * 0.1 * (1 - cos theta)

theta = 85.5 deg

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