A string is wrapped around the rim of a wheel of moment of inertia 0.20kgm2 and of radius 20cm . The wheel is free to rotate about its axis and initially it is at rest. The string is now pulled by a force of 20N . The angular velocity of the string after 5s will be

Question

A string is wrapped around the rim of a wheel of moment of inertia 0.20kgm2 and of radius 20cm . The wheel is free to rotate about its axis and initially it is at rest. The string is now pulled by a force of 20N . The angular velocity of the string after 5s will be <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-nllibnmlpo6qyiub.jpg" /> (A) 90rads- 1 (B) 70rads- 1 (C) 95rads- 1 (D) 100rads- 1

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-nllibnmlpo6qxfsf.jpg" /> We know that τ=Iα ...(.i.) and τ=rF=((20/100))(.20.) τ=4Nm Now from eq. (.i.) 4=(.0.2.)(.α .) So α =20rads- 2 So from ω =ω 0+Î±t ω 0=0andt=5sec. So ω =0+20(.5.) ω =100rad/sec.

Q. A string is wrapped around the rim of a wheel of moment of inertia $0.20 k g m^{2}$ and of radius $20 c m$ . The wheel is free to rotate about its axis and initially it is at rest. The string is now pulled by a force of $20 N$ . The angular velocity of the string after $5 s$ will be

$90 r a d s^{- 1}$

$70 r a d s^{- 1}$

$95 r a d s^{- 1}$

$100 r a d s^{- 1}$

We know that $τ = I α ... (i)$
and $τ = r F = (\frac{20}{100}) (20)$
$τ = 4 N m$
Now from eq. $(i)$
$4 = (0.2) (α)$
So $α = 20 r a d s^{- 2}$
So from $ω = ω_{0} + α t$
$ω_{0} = 0 an d t = 5 sec .$
So $ω = 0 + 20 (5)$
$ω = 100 r a d / sec .$

Q. A string is wrapped around the rim of a wheel of moment of inertia 0.20kgm2 and of radius 20cm . The wheel is free to rotate about its axis and initially it is at rest. The string is now pulled by a force of 20N . The angular velocity of the string after 5s will be

90rads−1

70rads−1

95rads−1

100rads−1