Question

A massless rod S having length 2 L has equal point masses attached to its two ends as shown in the fig.The rod is rotating about an axis passing through its centre and making angle alpha(α) with the axis The magnitude of the change of momentum of rod i.e, |dL/dt| equals:

a) 2 m l^3 ω^2 sinΦ•cisΦ
b) ml^2 ω^2 sin 2Φ
c) ml^2 sin2Φ
d) m ^1/2 l^1/2 ω sinΦ • cosΦ

[AnS WiTH ExPLaNATiON•••]​

Answer 1

Answer:

|dL/dt| = 2ml²ω²sin2θ

Explanation:

We have to find the magnitude of the change of momentum of rod |dL/dt|

so , As we know that angular momentum(L')

L = r×p = r×mv = rmωr [ ∵v= ωr]

∵ from given figure .

r = l sinα_______(1)

∴ | L|= l sinα(mω) l sinα

|L| = l²sin²α mω

• Now, |dL/dt| = d(I²sin²α mω|/dt

• |dL /dt| = l²mω 2sinα cosα (dα/dt )

Diffrentiating sin²α , we get 2sinαcosα (dα/dt)

• |dL/dt | = l² mω 2sinα cosα (dα/dt)

[∵ 2sinα cosα = sin2α ] and as we know that dα/dt = ω

∴|dL /dt | = mω² l² sin2α or mω²l²sin2α

But there is two equal mass attached with rod .

∴ 2mω²I² sin2θ Answer .

option (B) is correct .

Answer 2

|dL/dt| = 2ml²ω²sin2θ

Explanation:

We have to find the magnitude of the change of momentum of rod |dL/dt|

so , As we know that angular momentum(L')

L = r×p = r×mv = rmωr [ ∵v= ωr]

∵ from given figure .

r = l sinα_______(1)

∴ | L|= l sinα(mω) l sinα

|L| = l²sin²α mω

Now, |dL/dt| = d(I²sin²α mω|/dt

|dL /dt| = l²mω 2sinα cosα (dα/dt )

Diffrentiating sin²α , we get 2sinαcosα (dα/dt)

|dL/dt | = l² mω 2sinα cosα (dα/dt)

[∵ 2sinα cosα = sin2α ] and as we know that dα/dt = ω

∴|dL /dt | = mω² l² sin2α or mω²l²sin2α

But there is two equal mass attached with rod .

∴ 2mω²I² sin2θ Answer .

option (B) is correct .

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