SATHEE: Rotational Motion Question 139

Rotational Motion Question 139

Question: A uniform rod of length $L$ (in between the supports) and mass $m$ is placed on two supports $A$ and $B$ . The rod breaks suddenly at length $L / 10$ from the support B. Find the reaction at support A immediately after the rod breaks.

Options:

A) $\frac{9}{40} m g$

B) $\frac{19}{40} m g$

C) $\frac{m g}{2}$

D) $\frac{9}{20} m g$

Show Answer

Answer:

Correct Answer: A

Solution:

[a] Torque $= τ = \frac{9}{10} m g (\frac{9}{20} L) = I α = \frac{m}{3} {(\frac{9}{10} L)}^{2} α$

$α = \frac{3 g}{2 L}$

$Acceleration, α_{C M} = α (A C)$

$a_{C M} = \frac{3 g}{2 L} (\frac{9 L}{20}) = \frac{27 g}{40}$

$Now, \frac{9}{10} m g - N_{A} = m a_{C M} = m \cdot \frac{27 g}{40} Or$

$N_{A} = \frac{9}{40} m g$

Rotational Motion Question 139

Question: A uniform rod of length $L$ (in between the supports) and mass $m$ is placed on two supports $A$ and $B$ . The rod breaks suddenly at length $L / 10$ from the support B. Find the reaction at support A immediately after the rod breaks.

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Rotational Motion Question 139

Question: A uniform rod of length L (in between the supports) and mass m is placed on two supports A and B . The rod breaks suddenly at length L/10 from the support B. Find the reaction at support A immediately after the rod breaks.

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

Question: A uniform rod of length $L$ (in between the supports) and mass $m$ is placed on two supports $A$ and $B$ . The rod breaks suddenly at length $L / 10$ from the support B. Find the reaction at support A immediately after the rod breaks.