Practice Start Test

Rotational Motion

JEE Physics · 88 questions · Page 8 of 9 · Click an option or "Show Solution" to reveal answer

Q71
A bob of mass mm attached to an inextensible string of length ll is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed ω rad/s\omega \,rad/s about the vertical. About the point of suspension:
A angular momentum is conserved
B angular momentum changes in magnitude but not in direction.
C angular momentum changes in direction but not in magnitude.
D angular momentum changes both in direction and magnitude.
Correct Answer
Option C
Solution

Torque working on the bob of mass

mm

is,

τ=mg×ℓ sin⁡ θ.\tau = mg \times \ell \,\sin \,\theta .

(Direction of torque by the weight is parallel to plane of rotation of the particle) As τ\tau is perpendicular to the angular momentum

L→\overrightarrow L

of the bob, so the direction of

LL

changes but magnitude remains same.

Q72
Concrete mixture is made by mixing cement, stone and sand in a rotating cylindrical drum. If the drum rotates too fast, the ingredients remain stuck to the wall of the drum and proper mixing of ingredients does not take place. The maximum rotational speed of the drum in revolutions per minute(rpm) to ensure proper mixing is close to : (Take the radius of the drum to be 1.25 m and its axle to be horizontal) :
A 0.4
B 1.3
C 8.0
D 27.0
Correct Answer
Option D
Solution

For proper mixing, the concrete mixture should not stick to the top wall of the cylinder and it should fall down.

The maximum velocity at the top is v, then we have

mv2r=mg{{m{v^2}} \over r} = mg
⇒v=rg\Rightarrow v = \sqrt {rg}

Therefore, the maximum rotational speed of the drum is

ω=vr=gr=9.81.25\omega = {v \over r} = \sqrt {{g \over r}} = \sqrt {{{9.8} \over {1.25}}}

rad/s

=9.81.25×602π= \sqrt {{{9.8} \over {1.25}}} \times {{60} \over {2\pi }}

rpm = 26.74 ≈\approx 27 rpm

Q73
An electric dipole is formed by two equal and opposite charges q with separation d. The charges have same mass m. It is kept in a uniform electric field E. If it is slightly rotated from its equilibrium orientation, then its angular frequency ω\omega is :-
A qEmd\sqrt {{{qE} \over {md}}}
B qE2md\sqrt {{{qE} \over {2md}}}
C qE−2md\sqrt {{{qE} \over {-2md}}}
D 2qEmd\sqrt {{{2qE} \over {md}}}
Correct Answer
Option D
Solution

Moment of inertia

(I)=m(d2)2×2=md22(I) = m{\left( {{d \over 2}} \right)^2} \times 2 = {{m{d^2}} \over 2}

Now by

τ=lα\tau = l\alpha
(qE)(d sin⁡θ)=md22.α(qE)(d\,\sin \theta ) = {{m{d^2}} \over 2}.\alpha
α=(2qEmd)sin⁡θ\alpha = \left( {{{2qE} \over {md}}} \right)\sin \theta

for small θ\theta

⇒α=(2qEmd)θ\Rightarrow \alpha = \left( {{{2qE} \over {md}}} \right)\theta

⇒\Rightarrow Angular frequency

ω=2qEmd\omega = \sqrt {{{2qE} \over {md}}}
Q74
A thin smooth rod of length L and mass M is rotating freely with angular speed ω\omega 0 about an axis perpendicular to the rod and passing through its center. Two beads of mass m and negligible size are at the center of the rod initially. The beads are free to slide along the rod. The angular speed of the system , when the beads reach the opposite ends of the rod, will be :-
A Mω0M+3m{{M{\omega _0}} \over {M + 3m}}
B Mω0M+m{{M{\omega _0}} \over {M + m}}
C Mω0M+6m{{M{\omega _0}} \over {M + 6m}}
D Mω0M+2m{{M{\omega _0}} \over {M + 2m}}
Correct Answer
Option C
Solution

Initial angular momentum = Final Angular Momentum

ML212ω0=(ML212+2mL24)ω{{M{L^2}} \over {12}}{\omega _0} = \left( {{{M{L^2}} \over {12}} + 2{{m{L^2}} \over 4}} \right)\omega
⇒ω=Mω0M+6m\Rightarrow \omega = {{M{\omega _0}} \over {M + 6m}}
Q75
A thin disc of mass M and radius R has mass per unit area σ\sigma (r) = kr2 where r is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is :
A MR23{{M{R^2}} \over 3}
B MR26{{M{R^2}} \over 6}
C 2MR23{{2M{R^2}} \over 3}
D MR22{{M{R^2}} \over 2}
Correct Answer
Option C
Solution
IDisc=∫0R(dm)r2⇒IDisc=∫0R(σ2πrdr)r2{I_{Disc}} = \int\limits_0^R {\left( {dm} \right)} {r^2} \Rightarrow {I_{Disc}} = \int\limits_0^R {\left( {\sigma 2\pi rdr} \right)} {r^2}
IDisc=∫0R(kr22πrdr)r2{I_{Disc}} = \int\limits_0^R {\left( {k{r^2}2\pi rdr} \right)} {r^2}

Mass of Disc

IDisc=2πk∫0Rr2dr    M=∫0R2πrdr kr2{I_{Disc}} = 2\pi k\int\limits_0^R {{r^2}dr} \,\,\,\,M = \int\limits_0^R {2\pi rdr\,k{r^2}}
IDisc=2πk(r66)0R    M=2πk∫0Rr3dr{I_{Disc}} = 2\pi k\left( {{{{r^6}} \over 6}} \right)_0^R\,\,\,\,M = 2\pi k\int\limits_0^R {{r^3}dr}
IDisc=2πkR66    M=2πkr44∣0R{I_{Disc}} = 2\pi k{{{R^6}} \over 6} \,\,\,\,M = 2\pi k\left. {{{{r^4}} \over 4}} \right|_0^R
IDisc=πkR63=(πkR42)R223   M=2πkr44∣0R{I_{Disc}} = {{\pi k{R^6}} \over 3} = \left( {{{\pi k{R^4}} \over 2}} \right){{{R^2}2} \over 3}\,\,\,M = 2\pi k\left. {{{{r^4}} \over 4}} \right|_0^R
IDisc=M2R23;IDisc=23MR2{I_{Disc}} = {{M2{R^2}} \over 3};{I_{Disc}} = {2 \over 3}M{R^2}
Q76
A particle of mass m is moving along a trajectory given by x = x0 + a cosω\omega 1t y = y0 + b sinω\omega 2t The torque, acting on the particle about the origin, at t = 0 is :
A Zero
B +my0a ω12\omega _1^2k^\widehat k
C −m(x0bω22−y0aω12)k^ - m\left( {{x_0}b\omega _2^2 - {y_0}a\omega _1^2} \right)\widehat k
D m (–x0b + y0a) ω12\omega _1^2k^\widehat k
Correct Answer
Option B
Solution
F→=ma→=m[−aω12cos⁡ω,ti^−bω22sin⁡ω2tj^]\overrightarrow F = m\overrightarrow a = m\left[ { - a\omega _1^2\cos \omega ,t\widehat i - b\omega _2^2\sin {\omega _2}t\widehat j} \right]
f→t=0=−maω12i^{\overrightarrow f _{t = 0}} = - ma\omega _1^2\widehat i
r→t=0=(X0+a)i^+yj^{\overrightarrow r _{t = 0}} = \left( {{X_0} + a} \right)\widehat i + y\widehat j
τ→=r→×F→=my0aω12k^\overrightarrow \tau = \overrightarrow r \times \overrightarrow F = m{y_0}a\omega _1^2\widehat k
Q77
A person of mass M is, sitting on a swing of length L and swinging with an angular amplitude θ\theta 0. If the person stands up when the swing passes through its lowest point, the work done by him, assuming that his center of mass moves by a distance ℓ\ell (ℓ\ell << L), is close to;
A mgℓ\ell (1 + θ\theta 02)
B mgℓ\ell
C mgℓ\ell (1 + θ022{{\theta _0^2} \over 2})
D mgℓ\ell (1 - θ\theta 02)
Correct Answer
Option A
Solution

Angular momentum conservation MV0L = MV1(L –

ℓ\ell

)

V1=V0(LL−ℓ){V_1} = {V_0}\left( {{L \over {L - \ell }}} \right)
wg+wp=ΔKE{w_g} + {w_p} = \Delta KE
−mgℓ+wp=12m(V12−V02)- mg\ell + {w_p} = {1 \over 2}m\left( {V_1^2 - V_0^2} \right)
wp=mgℓ+12mV02((LL−ℓ)2−1){w_p} = mg\ell + {1 \over 2}mV_0^2\left( {{{\left( {{L \over {L - \ell }}} \right)}^2} - 1} \right)
=mgℓ+12mV02((1−LL−ℓ)−2−1)= mg\ell + {1 \over 2}mV_0^2\left( {{{\left( {1 - {L \over {L - \ell }}} \right)}^{ - 2}} - 1} \right)

Now

ℓ≪L\ell \ll L

By, Binomial approximation

=mgℓ+12mV02((1+LL−ℓ)−2−1)= mg\ell + {1 \over 2}mV_0^2\left( {{{\left( {1 + {L \over {L - \ell }}} \right)}^{ - 2}} - 1} \right)
=mgℓ+12mV02(2ℓL)= mg\ell + {1 \over 2}mV_0^2\left( {{{2\ell } \over L}} \right)
wp=mgℓ+mV02ℓL{w_p} = mg\ell + mV_0^2{\ell \over L}

Here, V0 = maximum velocity =

ω×A=(gL)(θ0L)\omega \times A = \left( {\sqrt {{g \over L}} } \right)\left( {{\theta _0}L} \right)

So,

wp=mgℓ+m(θ0gL)2ℓL{w_p} = mg\ell + m{\left( {{\theta _0}\sqrt {gL} } \right)^2}{\ell \over L}

=

mgℓ(1+θ02)mg\ell \left( {1 + \theta _0^2} \right)
Q78
A thin circular plate of mass M and radius R has its density varying as ρ\rho (r) = ρ\rho 0r with ρ\rho 0 as constant and r is the distance from its centre. The moment of Inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is I = aMR2. The value of the coefficient a is :
A 12{1 \over 2}
B 32{3 \over 2}
C 85{8 \over 5}
D 35{3 \over 5}
Correct Answer
Option C
Solution
M=∫0Rρ0r×2πrdr=2πρ0R33M = \int\limits_0^R {{\rho _0}r \times 2\pi rdr = {{2\pi {\rho _0}{R^3}} \over 3}}
IC=∫0Rρ0r×2πrdr×r2=2πρ0R53{I_C} = \int\limits_0^R {{\rho _0}r \times 2\pi rdr \times {r^2} = {{2\pi {\rho _0}{R^5}} \over 3}}

∴\therefore

I=IC+MR2=2πρ0R5(13+15)I = {I_C} + M{R^2} = 2\pi {\rho _0}{R^5}\left( {{1 \over 3} + {1 \over 5}} \right)

⇒\Rightarrow

16πρ0R515=85[23πρ0R3]R2=85MR2{{16\pi {\rho _0}{R^5}} \over {15}} = {8 \over 5}\left[ {{2 \over 3}\pi {\rho _0}{R^3}} \right]{R^2} = {8 \over 5}M{R^2}
Q79
To mop-clean a floor, a cleaning machine presses a circular mop of radius R vertically down with a total force F and rotates it with a constant angular speed about its axis. If the force F is distributed uniformly over the mop and if coefficient of friction between the mop and the floor is μ\mu , the torque, applied by the machine on the mop is -
A μ\mu FR/2
B μ\mu FR/3
C μ\mu FR/6
D 23{2 \over 3}μ\mu FR
Correct Answer
Option D
Solution

Consider a strip of radius x & thickness dx, Torque due to friction on this strip.

∫dτ=∫0RxμF.2πxdxπR2\int {d\tau = \int\limits_0^R {{{x\mu F.2\pi xdx} \over {\pi {R^2}}}} }
τ=2μFR2.R33\tau = {{2\mu F} \over {{R^2}}}.{{{R^3}} \over 3}
τ=2μFR3\tau = {{2\mu FR} \over 3}
Q80
The magnitude of torque on a particle of mass 1 kg is 2.5 Nm about the origin. If the force acting on it is 1 N, and the distance of the particle from the origin is 5m, the angle between the force and the position vector is (in radians) :
A π8{\pi \over 8}
B π6{\pi \over 6}
C π4{\pi \over 4}
D π3{\pi \over 3}
Correct Answer
Option B
Solution

2.5 = 1 ×\times 5 sin θ\theta sinθ\theta = 0.5 =

12{1 \over 2}

θ\theta =

π6{\pi \over 6}
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