SATHEE: Rotation 2 Question 6

Rotation 2 Question 6

6. A small mass $m$ is attached to a massless string whose other end is fixed at $P$ as shown in the figure. The mass is undergoing circular motion in the $x - y$ plane with centre at $O$ and constant angular speed $ω$ . If the angular momentum of the system, calculated about $O$ and $P$ are denoted by $L_{O}$ and $L_{P}$ respectively, then

(a) $L_{O}$ and $L_{P}$ do not vary with time

(b) $L_{O}$ varies with time while $L_{P}$ remains constant

(d) $L_{O}$ and $L_{P}$ both vary with time

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Answer:

Correct Answer: 6. (c)

Solution:

Angular momentum of a particle about a point is given by $L = r \times p = m (r \times v)$

For $L_{O}$

$| L | = (m v r \sin θ) = m (R ω) (R) \sin 90^{\circ} =$ constant Direction of $L_{O}$ is always upwards. Therefore, complete $L_{O}$ is constant, both in magnitude as well as direction.

For $L_{P}$

$| L_{P} | = (m v r \sin θ) = (m) (R ω) (l) \sin 90^{\circ} = (m R l ω)$

Magnitude of $L_{P}$ will remain constant but direction of $L_{P}$ keeps on changing.

Rotation 2 Question 6

Answer:

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Rotation 2 Question 6

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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