First slide

Rotational motion

Question

A block of mass m = 1 kg slides with velocity $υ = 6 m / s$ on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about O and swings as a result of the collision making angle $θ$ before momentarily coming to rest. If the rod has mass M = 2 kg, and length l = 1 m, the value of $θ$ is approximately: $(t a k e g = 10 m / s^{2})$

Difficult

A

69º
B

55º
C

63º
D

49º

Solution

Initial angular momentum of block about hinge before Collison = Final angular momentum of combine system after collision
$\Rightarrow m v l = L - - - (1)$
Now, using energy conservation
$\begin{array}{l} L o s s i n r o t a t i o n a l k i n e t i c e n e r g y o f c o m b i n e s y s t e m = G a i n i n p o t e n t i a l e n e r g y o f c o m b i n e s y s t e m \\ \Rightarrow \frac{L^{2}}{2 I} = m g l (1 - \cos θ) + M g \frac{l}{2} (1 - \cos θ) (∵ \cos θ = \frac{l - h}{l} \Rightarrow h = l (1 - \cos θ)) \end{array}$
$\begin{array}{l} \Rightarrow \frac{{(m v l)}^{2}}{2 (m l^{2} + \frac{M l^{2}}{3})} = (m + \frac{M}{2}) g l (1 - \cos θ) \\ (∵ I = I_{b l o c k} + I_{r o d} = m l^{2} + \frac{M l^{2}}{3} & f r o m (1)) \\ \Rightarrow \frac{{(1 \times 6 \times 1)}^{2}}{2 (1 \times 1^{2} + \frac{2 \times 1^{2}}{3})} = (1 + \frac{2}{2}) 10 \times 1 (1 - \cos θ) \\ \Rightarrow \cos θ = \frac{23}{50} \\ (∵ m = 1 k g, l = 1 m a n d M = 2 k g) \\ \Rightarrow θ = 63^{0} \end{array}$

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