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Rotation


Question 1. A mass is whirled in a circular path with constant angular velocity and its angular momentum is L.If the string is now halve keeping the angular velocity same then angular momentum is
a. L
b. L/4
c. L/2
d. 2L

Solution 1
Angular momentum for this is defined as
=mr2ω

First case
L=mr2ω

Second case
Lf=m(r/2)2ω

So Lf=L/4

Question 2.A mass is moving with constant velocity along a line parallel to xaxis away from origin.its angular momentum with respect to origin is
a. is zero
b. remains constant
c. goes on increasing
d. goes on decreasing

Solution 2

L=(mv)Xr
or
L=mvrsinθ
Now rsinθ=perpendicular distance from x axis which is constant
So Angular momentum is constant

Question 3.A cylinder rolls up the incline plane reaches some height and then roll down without slipping through out this section.The direction of the frictional force acting on the cylinder are
a. Up the incline while ascending and down the incline while descending
b.Up the incline while ascending and desending
c. down the incline while ascending and up the incline while descending
d.down the incline while ascending and desending
Solution 3:
Imagine the cylinder to be moving on a frictionless surface.In both the cases the acceleration of the CM of the cylinder is gsinθ.This is also the acceleration of the point of contact
of the cylinder and the inclined plane..Also no torque (about the center of the cylinder) is acting on the cylinder since we assumed the surface to be a frictionless and the forces
acting on the cylinder is mg and N which passes through the center of the cylinder.Therefore the net movement of the point of contact in both the cases is in downward direction
Therefore frictional force will act in upward direction in both the cases

Question 4.A uniform sold sphere rolls on the horizontal surface at 20 m/s.it then rolls up the incline of 30.If friction losses are negligible what will be the value of h where sphere stops on the incline
a. 28.6 m
b 30 m
c. 28 m
d. none of these

Solution 4

Let h be the height

The rotational and translational KE of the ball at the bottom will be changed to Gravitational energy when the sphere stops .
We therefore writes

(1/2)Mv2+(1/2)Iω2=Mgh

For a solid sphere I=(2/5)Mr2 and also ω=v/r

So (1/2)Mv2+ (1/2)(2/5)Mr2(v/r)2=Mgh
or
v2/2 +v2/5=gh

or h=28.6 m


Question 5. A cylinder of Mass M and radius R rolls down a incline plane of inclination θ.Find the linear accleration of the cylinder
a. (2/3)gsinθ
b.(2/3)gcosθ
c gsinθ
d none of these

Solution 5

Net force on the cylinder
Fnet=mgsinθ -f
or ma=mgsinθ -f
Where f is the frictional force
Now τ=fXR=Iα
Now in case of pure rolling we know that
a=αR => α=a/R

So f=Ia/R2

From 1 and 2
a=mgsinθ /[m+(I/R2)]

Now I=mR2/2

So a=(2/3)gsinθ



Question 6 An ice skater spins with arms outstretch at 1.9 rev/s.Her moment of inertia at this time is 1.33 kgm2.She pulls her arms to increase her rate of spin.Her moment of inertia after she pulls her arm is .48kgm2.What is her new rate of spinning
a. 5.26 rev/s
b. 5.2 rev/s
c 4.7 rev/s
d. none of thes

Solution 6

Law of conservation of angular momentum

I1ω1=I2ω2
or

1.33(1.9)=.48ω2
or
ω2=5.26 rev/s


Question 7. Moment of inertia of a uniform rod of lenght L and mass M about an axis passing through L/4 from one end and perpendicular to its lenght
a. 7ML2/36
b.7ML2/48
c. 11ML2/48
d.ML2/12

Solution 7

Using parallel axis theorem
I=Icm+Mx2 where x is the distance of the axis of the rotation from the CM of the rod
So x=L/2-L/4=L/4 Also Icm=ML2 /12

So I=ML2 /12+ML2 /16=7ML2 /48


Question 8. A wheel starts from rest and spins with a constant angular acceleration. As time goes on the
acceleration vector for a point on the rim:

a. increases in magnitude but retains the same angle with the tangent to the rim

b.increases in magnitude and becomes more nearly radial

c. increases in magnitude and becomes more nearly tangent to the rim

d. decreases in magnitude and becomes more nearly radial

Solution 8

Tangential acceleration=radius* angular acceleration
Since angular acceleration is constant ...Tangential acceleration is constant

Radial acceleration=r* (angular velocity)2
Since angular velocity increase with time...Radial acceleration increase with time

So resulttant acceleration increase with time and becomes more radial as time passes

Question 9.
Two wheels are identical but wheel B is spinning with twice the angular speed of wheel A. The ratio of the
magnitude of the &radical acceleration of a point on the rim of B to the magnitude of the radial acceleration of
a point on the rim of A is:

a. 4
b . 2
c 1/2
d 1/4

Solution 9.
Radial acceleration=r* (ω)2

For wheel A
Radial acceleration of A =r* (ω)2

For wheel B
Radial acceleration of B=r* (2ω)2=4r* (ω)2

So Radial acceleration of B/Radial acceleration of A=4:1

Question 10. For a wheel spinning with constant angular acceleration on an axis through its center, the ratio of
the speed of a point on the rim to the speed of a point halfway between the center and the rim is:

a 2
b 1/2
c 4
d 1/4

Solution 10

At rim

v=rω

At point between the center and rim
v=(r/2)ω

Ratio =2


Question 11. A wheel initially has an angular velocity of 18 rad/s. It has a constant angular acceleration of 2.0 rad/s2 and is
slowing at first. What time elapses before its angular velocity is18 rad/s in the direction opposite to its initial
angular velocity?

a 3 sec
b 6 sec
c 18 sec
d none of these

Solution 11

ω0=18
ω=-18

anugular acceleration(α)=-2

Now

ω=ω0+αt
or t=18


Question 12. One solid sphere X and another hollow sphere Y are of same mass and same outer radii. Their moment of inertia about their diameters are respectively Ix and Iy such that
(A) Ix= Iy     
(B) Ix > Iy
(C) Ix < Iy
(D) Ix/Iy=Dx/Dy
          Where Dx and Dy are their densities.


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