Q: A wheel of radius 20 cm can rotate freely about its centre as shown in the figure. When a light string is wrapped over the rim and it is pulled down with a force of 5 N, it is found that an angular acceleration of 2rads−2 is produced on the wheel. Find the moment of inertia of the wheel about its axis of rotation.

Torque τ=∣r×F∣=5×0.2=1 N m Also torque τ=Iα ∴I=τ/α=21=0.5 kg m2

Q: A cylinder of mass M is suspended by means of two light, inextensible strings wrapped around it as shown. Find the tension in the string and the velocity of the cylinder as it falls through a height h.

mg−2 T=ma2Tr=21mr2α=21mra or 2 T=21ma Solving (1) and (2) a=32 g and T=6mg v2=2[32 g]h(∴v2=2ah)∴v=34gh

Q: A uniform sphere of mass m, and radius R starts rolling without slipping down an inclined plane of angle θ after a time t from start. Find the angular momentum of the sphere relative to the initial point of contact, after time t.

mgsinθ−Ff=ma FfR=Iα=RIa mgsinθ=a(m+R2I)=a(1+52)m a=75 gsinθ ω=Rv=R at ⇒v=at L=mvR+ICMω=mRat+ICMR at =maRt(1+52)=maRt57=mgsinθ⋅Rt75⋅57 L=mgRsinθ.t

Q: A cylinder is released from the top of an incline of inclination θ with horizontal and length ℓ. If the cylinder rolls without slipping, find the speed with which it reaches the bottom.

Gain in KE= loss in PE 21Iω2+21mv2=mgℓsinθ⇒21[21mR2]ω2+21mv2=43mv2=mgℓsinθ∴v=34 gℓsinθ

Question 1

A wheel of radius 20 cm can rotate freely about its centre as shown in the figure. When a light string is wrapped over the rim and it is pulled down with a force of 5 N, it is found that an angular acceleration of 2rads−2 is produced on the wheel. Find the moment of inertia of the wheel about its axis of rotation.

Accepted Answer

Torque τ=∣r×F∣=5×0.2=1 N m
Also torque τ=Iα
∴I=τ/α=21​=0.5 kg m2

Question 2

A cylinder of mass M is suspended by means of two light, inextensible strings wrapped around it as shown. Find the tension in the string and the velocity of the cylinder as it falls through a height h.

Accepted Answer

mg−2 T=ma2Tr=21​mr2α=21​mra​
or 2 T=21​ma
Solving (1) and (2)
a=32​ g and T=6mg​
v2=2[32​ g]h(∴v2=2ah)∴v=34gh​​​

Question 3

A thin ring of mass m has a long, thin, inextensible string of mass m ' wrapped around it. If the system is released as shown, find its linear acceleration.

Accepted Answer

Let ℓ be the length, λ the linear density and η the fraction of length remaining on the ring.
 Mass =m+ηλℓ⇒(m+ηλℓ)g−T=(m+ηλℓ)aTr=Iα=(m+ηλℓ)r2ra​∴T=(m+ηλℓ)a substitute (2) in equation (1)(m+ηλℓ)g−(m+ηλℓ)a=(m+ηλℓ)aa=2g​, vertically downwards. ​

Question 4

Two masses of 0.5 kg and 0.8 kg are connected by an inextensible light thread passing over a pulley of mass 1 kg and radius 0.1 m. Calculate (no slipping between pulley and the thread) (g=10 m s−2)
(i) the acceleration of the masses and the pulley.
(ii) the tensions in the vertical portions of the string, if it does not slip.

Accepted Answer

(i) Let the two masses connected by an inextensible light thread be m1​ and m2​. Equations of motion,
T1​−m1​ g=m1​am2​ g−T2​=m2​a​
Considering rotational motion of the pulley (a disc),
τ=T2​R−T1​R=21​MR2×α(∴τ=Iα)
where α= angular acceleration of the pulley. But a=αR for no slip
∴(T2​−T1​)Ra=​=21​MR2×Ra​⇒T2​−T1​=21​ Ma Solving (1), (2) and (3), m1​+m2​+2M​(m2​−m1​)g​=1.8+0.300×10​=6+10​ m s−2​
(ii) m2​=800 gram, m1​=500gram,M=1 kg
From equation
T1​=m1​( g+a)=0.5×(10+610​)=5.83 N
From equation
T2​=m2​( g−a)=0.8×650​=6.67 N

Question 5

A uniform sphere of mass m, and radius R starts rolling without slipping down an inclined plane of angle θ after a time t from start. Find the angular momentum of the sphere relative to the initial point of contact, after time t.

Accepted Answer

mgsinθ−Ff​=ma
Ff​R=Iα=RIa​
mgsinθ=a(m+R2I​)=a(1+52​)m
a=75​ gsinθ
ω=Rv​=R at ​⇒v=at
L=mvR+ICM​ω=mRat+ICM​R at ​=maRt(1+52​)=maRt57​=mgsinθ⋅Rt75​⋅57​
 L=mgRsinθ.t

Question 6

A thin ring, of mass 2 m and radius r, in the horizontal plane can rotate freely about a vertical axis through its centre. A body of mass m is placed inside the ring and given a velocity v0​ as shown. If the inside of the ring has coefficient of friction μ, find the final angular velocity.

Accepted Answer

The angular momentum about the centre of the ring is conserved. Initial angular momentum =mv0​⋅r
Let the final angular velocity be ω and the final velocity of m be v⇒v=ωr Final angular momentum =Iω+mvr=2 m⋅r2.ω+mvr=mvr(2+1)=3mvr=mv0​r ⇒v=3v0​​⇒ω=rv​=3rv0​​

Question 7

A bullet of mass m travelling with a speed v hits a disc of mass M and radius R as shown and gets stuck to it. Find the angular velocity of the system.

Accepted Answer

By conservation of angular momentum about ' C ' mvsinα⋅R=(m+2M​)R2.ω
ω=(2m+M2m​)Rvsinα​

Question 8

A uniform circular horizontal rough platform, free to rotate about a vertical axis through its centre is at rest along with a man whose mass is 41​ th of that of platform. The man walks round the edge completing one round and comes back to his starting point. Determine the angle through which the platform has turned.

Accepted Answer

L about axis through O is zero and conserved.
Let mass of man=m and mass of platform =4 m.
Radius =R
At any instant let ωrel ​ be angular velocity of man relative to platform and let ω be angular velocity of platform.
Then, 42mR2​ω+mR2(ω+ωrel​)=0
⇒ω=3−ωrel​​⇒dtdθ​=−3dtdθrel​​⇒dθ=−3dθrel​​⇒θ=−θrel​/3(∵ at θ=0,θrel​=0)​
Putting θrel ​=2π⇒θ=3−2π​
(negative sign shows platform rotates in the opposite direction)

Question 9

A uniform circular disc of mass 100 g and radius, 2 cm is rotated about one of its diameters at an angular speed of 10rads−1. Find the kinetic energy of the disc and angular momentum about the axis of rotation.

Accepted Answer

Moment of inertia of the disc about one of its diameters =41​MR2 =41​×0.1 kg×(0.02)2=1×10−5 kg m2
K.E. =21​Iω2=21​×1×10−5×100=5×10−4 J
Angular momentum about the axis of rotation =L=Iω=1×10−5×10rads−2 =1×10−4Js

Question 10

A cylinder is released from the top of an incline of inclination θ with horizontal and length ℓ. If the cylinder rolls without slipping, find the speed with which it reaches the bottom.

Accepted Answer

Gain in KE= loss in PE
21​Iω2+21​mv2=mgℓsinθ⇒21​[21​mR2]ω2+21​mv2=43​mv2=mgℓsinθ∴v=34​ gℓsinθ​​

Question 11

A uniform rod of length 1 m and mass 0.5 kg is held with one end resting on a smooth horizontal table making 30∘ with the vertical. Find its angular velocity when the rod is released and makes 60∘ with the vertical. Calculate the angular momentum about the center of mass.

Accepted Answer

C.M of the rod will fall vertically. The rod has rotational motion about the vertical velocity axis. Let v be the vertical velocity of the C.M of the rod when it is inclined at θ with the vertical. Let α be the initial angle and length 2 a. From conservation of energy
mgacosα=mgacosθ+21​ICM​ω2+21​mv2
But I=121​ m(2a)2=31​ma2 and v=v′cos(90∘−θ)
=v′sinθ=ωasinθ∴mgacosα=mgacosθ+21​(31​ma2)ω2+21​ mω2a2sin2θ​
or ga cosα=gacosθ+61​a2ω2+21​a2ω2sin2θ
ga(cosα−cosθ)=a2ω2(61​+21​sin2θ)⇒[a(1+3sin2θ)6g(cosα−cosθ)​]21​=ω
2a=1 m,∴a=0.5 m;g=10 m s−2;α=30∘,θ=60∘⇒ω=(0.5(1+3×0.75)60(0.866−0.5)​)21​
ω=3.68rads−1
Angular momentum of the rod =ICM​ω=31​ma2ω=31​×0.5×0.25×3.68 =0.153 kg m2 s−1

Question 12

A uniform round object with radius R and radius of gyration K, is placed on an inclined plane of inclination θ to horizontal. What should be the minimum coefficient of friction so that the object rolls (without slipping)?

Accepted Answer

Condition for pure rolling:
Rω=v⇒R(ω0​+αt)=v0​+ at ⇒Rα=a(∵ω0​ and v0​ are zero )mgsinθ−f=ma f. R=Iα=m2α (2) ⇒f=RmK2α​​

Substitute in (1)
⇒a=mmgsinθ−f​=gsinθ−RK2α​
substituting condition for rolling viz. Rα=a
α(1+R2K2​)=Rgsinθ​⇒α=R(1+R2K2​)gsinθ​
substituting in (2)
f=R2mk2​⋅(1+R2K2​)gsinθ​
But f≤fmax​=μmgcosθ
⇒R2mK2​⋅(1+R2K2​)gsinθ​≤μmgcosθ⇒μ≥(1+ K2R2​)tanθ​
∴μmin​=(1+ K2R2​)tanθ​

Question 13

A uniform round object with radius R and radius of gyration K, is placed on an inclined plane of inclination θ to horizontal. What should be the minimum coefficient of friction so that the object rolls (without slipping)?

Accepted Answer

Condition for pure rolling:
Rω=v⇒R(ω0​+αt)=v0​+ at ⇒Rα=a(∵ω0​ and v0​ are zero )mgsinθ−f=ma f. R=Iα=m2α (2) ⇒f=RmK2α​​

Substitute in (1)
⇒a=mmgsinθ−f​=gsinθ−RK2α​
substituting condition for rolling viz. Rα=a
α(1+R2K2​)=Rgsinθ​⇒α=R(1+R2K2​)gsinθ​
substituting in (2)
f=R2mk2​⋅(1+R2K2​)gsinθ​
But f≤fmax​=μmgcosθ
⇒R2mK2​⋅(1+R2K2​)gsinθ​≤μmgcosθ⇒μ≥(1+ K2R2​)tanθ​
∴μmin​=(1+ K2R2​)tanθ​

Question 14

A uniform rod of mass M and length L is lying on a smooth table. A small particle of mass m moving on the surface of the table with velocity v0​ hits the rod normally at a distance d from the centre. The particle m comes to rest after a perfect elastic collision.
(i) Find an expression for ratio Mm​
(ii) If the collision is at one end of the rod, what is the value of ω and the velocity of the end point?

Accepted Answer

(i) Assuming m comes to rest after collision, using conservation of linear momentum:
mv0​=Mv⇒v=Mmv0​​
From conservation of angular momentum:
mv0​ d=12ML2​ω⇒ω=mv0​ML212 d​
conservation of energy: 21​Mv2+21​Iω2=21​mv02​
Substitute from (1) and (2)
⇒1=Mm​=​21​mv0​2=21​MM2m2v0​2​+21​12ML2​M2 L4m2v0​2(12)2 d2​Mm​+Mm​ L212 d2​ L2+12 d2L2​​
(ii) From equations (1) and (2) above,
ω=Mm​ L2v0​12 d​=41​ L2v0​122L​​=2 L3v0​​(∴d=2L​,Mm​=41​)
velocity of end point:
v′=v+ω⋅2L​=Mm​v0​+2L​2L3v0​​=4v0​​+43v0​​=v0​= velocity of approach (elastic collision)

Question 15

A body of mass m and speed v hits a rod of length ℓ and mass m at a point 3ℓ​ from its fixed support as shown and sticks to it. Find the angular velocity of the rod.

Accepted Answer

Angular momentum of system is conserved about the fixed axis of rotation, as the net external torque on system is zero.
∴=⇒ω=​Li​=Lf​mvr1​=mr1​v1​+I0​ω(v1​=r1​ω)mr12​ω+31​ mℓ2ω3mvℓ​=(m(3ℓ​)2+31​ mℓ2)ω(31​+91​)ℓv/3​=4ℓ3v​​

Question 16

Two uniform spherical balls of masses 20 kg,10 kg and radii 0.2 m,0.1 m respectively, are attached to the ends of a rod of mass 10 kg, and length 1 m. The balls and rod are homogenous and of uniform composition. Determine the moment of inertia of this system about an axis passing through its centre of mass and perpendicular to the rod.

Accepted Answer

With origin at mid point of rod,
xCM​==ICM​=⇒+52​ m2​r22​+m2​(r2​+2ℓ​+xCM​)2ICM​=+52​×10×(0.1)2+10×(0.1+0.5+0.2)2+10×121​+10(0.2)2=13 kg m2​M+m1​+m2​−m1​(r1​+2ℓ​)+m2​(r2​+2ℓ​)​10+20+1020(−0.2−0.5)+10(0.1+0.5)​=−0.2 m52​ m1​r12​+m1​(r1​+2ℓ​−xCM​)2+12Mℓ2​+M(xCM​)252​×20(0.2)2+20(0.2+0.5−0.2)2​

Question 17

Determine moment of inertia of a solid cylinder of mass M, radius R, length ℓ about an axis YY passing through its centre of mass and perpendicular to the axis of cylinder.

Accepted Answer

Consider thin disc, dz thick, radius R at z from YY axis. Moment of inertia of this thin disc about its axis parallel to YY is
dI=dm4R2​
dm=πR2ρdz (where ρ=πR2ℓM​= density of cylinder
material); dIyy​=dI+dmz2
For solid cylinder,
Iyy​==4πR4ρ​∫−ℓ/2+ℓ/2​dz+πR2ρ∫−ℓ/2+ℓ/2​z2dzρ.πR2ℓ[4R2​+12ℓ2​]​∫−ℓ/2+ℓ/2​dm4R2​+∫−ℓ/2+ℓ/2​dmz=M(4R2​+12ℓ2​)=12M​(3R2+ℓ2)​

Filters

Filters

book

topic

difficulty

Solutions for IIT JEE Super Course in Physics Mechanics II by Trishna Knowledge Systems

Found the solution, but did not understand the concept?

Ask my first question