MHT CET · Physics · Rotational Motion

A mass is whirled in a circular path with constant angular velocity and its linear velocity is v. If the string is now halved keeping the angular momentum same, the linear velocity is

A 2v
B $\frac{v}{2}$
C $v$
D $v \sqrt{2}$

Verified Solution

Answer & Solution

Correct Answer

(A) 2v

Step-by-step Solution

Detailed explanation

Angular velocity =

ω

Linear velocity =

v

Length of string = Radius =

r

If string is halved,

r^{'} = \frac{r}{2}

Angular momentum,

L = r \times p

where,

r

= radius vector
and p = linear momentum

= r \times m v [∵ θ = 90^{o}]

= m v r

As L remain constant,

m v r

= constant

\Rightarrow

r^{'} \to \frac{r}{2^{'}} v^{'} = 2 v

such that

m v^{'} r^{'} = m 2 v \times \frac{r}{2}

Hence, linear velocity will be 2 v.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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