SATHEE: Rotation 4 Question 7

Rotation 4 Question 7

8. Two uniform rods $A$ and $B$ of length $0.6 m$ each and of masses $0.01 k g$ and $0.02 k g$ , respectively are rigidly joined end to end. The combination is pivoted at the lighter end, $P$ as shown in figure. Such that it can freely rotate about point $P$ in a vertical plane.

A small object of mass $0.05 k g$ , moving horizontally, hits the lower end of the combination and sticks to it. What should be the velocity of the object, so that the system

could just be raised to the horizontal position?

(1994, 6M)

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Answer:

Correct Answer: 8. $6.3 m / s$

Solution:

System is free to rotate but not free to translate. During collision, net torque on the system ( $rod A + rod B + mass m)$ about point $P$ is zero.

Therefore, angular momentum of system before collision

$=$ angular momentum of system just after collision (about $P$ ).

Let $ω$ be the angular velocity of system just after collision, then

$\Rightarrow m v (2 l) = I ω$

$L_{i} = L_{f}$

Here, $I =$ moment of inertia of system about $P$

$= m (2 l)^{2} + m_{A} (l^{2} / 3) + m_{B} \frac{l^{2}}{12} + \frac{l}{2} + l^{2}$

Given, $l = 0.6 m, m = 0.05 k g, m_{A}$

$= 0.01 k g and m_{B} = 0.02 k g .$

Substituting the values, we get

Therefore, from Eq. (i)

$\begin{aligned} ω & = \frac{2 m v l}{I} = \frac{(2) (0.05) (v) (0.6)}{0.09} \\ ω & = 0.67 v \end{aligned}$

Now, after collision, mechanical energy will be conserved. Therefore, decrease in rotational KE

$=$ increase in gravitational $P E$ or $\frac{1}{2} I ω^{2} = m g (2 l) + m_{A} g \frac{l}{2} + m_{B} g (l + \frac{l}{2})$

or $ω^{2} = \frac{g l (4 m + m_{A} + 3 m_{B})}{I}$

$\begin{aligned} = \frac{(9.8) (0.6) (4 \times 0.05 + 0.01 + 3 \times 0.02)}{0.09} \\ = 17.64 (r a d / s)^{2} \end{aligned}$

$∴ ω = 4.2 r a d / s$

Equating Eqs. (ii) and (iii), we get

$v = \frac{4.2}{0.67} m / s or v = 6.3 m / s$

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Rotation 4 Question 7

8. Two uniform rods $A$ and $B$ of length $0.6 m$ each and of masses $0.01 k g$ and $0.02 k g$ , respectively are rigidly joined end to end. The combination is pivoted at the lighter end, $P$ as shown in figure. Such that it can freely rotate about point $P$ in a vertical plane.

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Rotation 4 Question 7

8. Two uniform rods A and B of length 0.6m each and of masses 0.01kg and 0.02kg, respectively are rigidly joined end to end. The combination is pivoted at the lighter end, P as shown in figure. Such that it can freely rotate about point P in a vertical plane.

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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8. Two uniform rods $A$ and $B$ of length $0.6 m$ each and of masses $0.01 k g$ and $0.02 k g$ , respectively are rigidly joined end to end. The combination is pivoted at the lighter end, $P$ as shown in figure. Such that it can freely rotate about point $P$ in a vertical plane.