**In the above problem the angular velocity of the system after the particle sticks to it will be**

__Question:__**Options:**

A. | 0.3 rad/s | |

B. | 5.3 rad/s | |

C. | 10.3 rad/s | |

D. | 89.3 rad/s |

**Answer: Option C**

: C

Initial angular momentum of bullet + initial angular momentum of cylinder

= Final angular momentum of (bullet + cylinder) system

⇒mvr+I1ω=(I1+I2)ω′

⇒mvr+I1ω=(12Mr2+mr2)ω′

⇒0.5×5×0.2+0.12=(122(0.2)2+(0.5)(0.2)2)ω′

∴ω′=10.3rad/sec.

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## More Questions on This Topic :

**A cord is wound round the circumference of wheel of radius 'r'. The axis of the wheel is horizontal and moment of inertia about it is 'I'. A weight 'mg' is attached to the end of the cord and falls from rest. After falling through a distance h, the angular velocity of the wheel will be**

__Question 1.__- √2ghI+mr
- √2mghI+mr2
- √2mghI+2mr2
- √2gh

**Answer: Option B**

: B

According to law of conservation of energy mgh=12(I+mr2)ω2⇒ω=√2mghI+mr2.

**A cylinder is released from rest from the top of an inclined plane of inclination θ and length 'l'. If the cylinder rolls without slipping, what will be its speed when it reaches the bottom?**

__Question 2.__- √34gl cosθ
- √34gl sinθ
- √34gl tanθ
- √34gl cotθ

**Answer: Option B**

: B

Let the mass of the cylinder be m and its radius r. Suppose the linear speed of the cylinder when it reaches the bottom is v. As the cylinder rolls without slipping, its angular speed about its axis is ω=vr. The kinetic energy at the bottom will be= 34mv2

Note: Try reducing this result on your own for practice. Thus, 34mv2=mglsinθ or, v=√43glsinθ

**A particle is projected at time t =0 from a point P with a speed v0 at an angle of 45∘ to the horizontal. Find the magnitude and the direction of the angular momentum of the particle about the point P at time t =v0g.**

__Question 3.__- mv302√2g^j
- mv302√2g^(−j)
- mv30√2g^j
- 2mv303^(−j)

**Answer: Option B**

: B

Let us take the origin at P, X-axis along the horizontal and Y-axis along the vertically upwards direction as shown in figure. For the horizontal motion during the time 0 to t, vx=v0cos45∘=v0√2 and x=vxt=v0√2.v0g=v20√2g For vertical motion, vγ=v0sin45∘−gt=v0√2−v0=1−√2√2v0 and y=(v0sin45∘)t−12gt2 y=v20√2g−v202g=v202g(√2−1) The angular momentum of the particle at time t about the origin is L=⃗r×⃗p=m⃗r×⃗v =m(→ix+→jy)×(−→ivx+−→jvy) =m(⃗kxvy−→kyvx) =m⃗k[(v20√2g)v0√2(1−√2)−v202g(√2−1)v0√2] =−⃗kmv302√2g Thus, the angular momentum of the particle is mv302√2g in the negative Z-direction, i.e., perpendicular to the plane of motion, going into the plane.

**A solid sphere of mass 0.1 kg and radius 2 cm rolls down an inclined plane 1.4m in length at an angle whose slope is 1/10. If sphere started from rest, its final velocity will be**

__Question 4.__- 1.4 m/sec
- 0.14 m/sec
- 14 m/sec
- 0.7 m/sec

**Answer: Option A**

: A

v=√2gh1+k2R2=√2×9.8×lsinθ1+25[Ask2R2=25,l=hsinθandsinθ=110given]

⇒v=√2×9.8×1.4×11075=1.4m/s.

**A smooth uniform rod of length 'L' and mass 'M' has two identical beads of negligible size, each of mass 'm', which can slide freely along the rod. Initially the two beads are at the centre of the rod and the system is rotating with angular velocity ω0 about an axis perpendicular to the rod and passing through the mid point of the rod (see figure). There are no external forces. When the beads reach the ends of the rod, the angular velocity of the system is**

__Question 5.__- ω0
- Mω0M+12m
- Mω0M+2m
- Mω0M+6m

**Answer: Option D**

: D

Since there are no external forces therefore the angular momentum of the system remains constant.

Initially when the beads are at the centre of the rod angular momentum L1=(ML212)ω0 .....(i)

When beads reach the ends of the rod then angular momentum =(m(L2)2+m(L2)2+ML212)ω′

..(ii) Equating (i) and (ii) ML212ω0=(mL22+ML212)ω′⇒ω′=Mω0M+6m.

**A smooth sphere A is moving on a frictionless horizontal plane with angular speed ω and centre of mass is moving translationally with velocity 'v'. It collides elastically and head-on with an identical sphere B which is at rest. All surfaces are frictionless. After the collision, their angular speeds are ωA and ωB, respectively. Then,**

__Question 6.__- ωB=0
- ωA=ωB
- ωA
- ωB=ω

**Answer: Option A**

: A

ωB=0rotational KE can’t be transferred from A to B as surfaces are frictionless.

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