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A thin circular ring of mass ‘M and radius r is rotating about its axis with a constant angular velocity ω, Two objects, each of mass m,, are attached gently to the opposite ends of a diameter of the ring. The wheel now rotates with an angular velocity
Since the objects are placed gently, therefore no external torque is acting on the system. Hence angular momentum is constant.
i.e., I_{1}w_{1} = I_{2} w_{2}
Two point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod of length 1.4 m and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of
The moment of inertia of the system about axis of rotation O is
I = I_{1} + I_{2} = 0.3x^{2} + 0.7 (1.4 – x)^{2} = 0.3x^{2} + 0.7 (1.96 + x^{2} – 2.8x)
= x^{2} + 1.372 – 1.96x
The work done in rotating the rod is converted into its rotational kinetic energy.
For work done to be minimum
A smooth sphere A is moving on a frictionless horizontal plane with angular speed ω and centre of mass velocity ν. It collides elastically and head on with an identical sphere B at rest. Neglect friction everywhere. After the collision, their angular speeds are ω_{A} and ω_{B}, respectively. Then
As the spheres are smooth there will be no friction (no torque) and therefore there will be no transfer of angular momentum. Thus A, after collision will remain with its initial angular momentum. i.e., ω_{ A} =ω
A disc of mass M and radius R is rolling with angular speed ω on a horizontal plane as shown in Figure. The magnitude of angular momentum of the disc about the origin O is
KEYCONCEPT. The disc has two types of motion namely translational and rotational. Therefore there are two types of angular momentum and the total angular momentum is the vector sum of these two.
In this case both the angular momentum have the same direction (perpendicular to the plane of paper and away from the reader).
LT = angular momentum due to translational motion.
L_{R} = angular momentum due to rotational motion about C.M.
L = MV × R + I_{cm}ω
I_{cm} = M.I. about centre of mass C.
(v = Rω in case of rolling motion and surface at rest)
A cubical block of side a is moving with velocity V on a horizontal smooth plane as shown in Figure. It hits a ridge at point O. The angular speed of the block after it hits O is
Net torque about O is zero.
Therefore, angular momentum (L) about O will be conserved, or L_{i }= L_{f}
A long horizontal rod has a bead which can slide along its length and initially placed at a distance L from one end A of the rod. The rod is set in angular motion about A with constant angular acceleration a . If the coefficient of friction between the rod and the bead is m , and gravity is neglected, then the time after which the bead starts slipping is
Note : When we are giving an angular acceleration to the rod, the bead is also having an instantaneous acceleration a = Lα. This will happen when a force is exerted on the bead by the rod. The bead has a tendency to move away from the centre. But due to the friction between the bead and the rod, this does not happen to the extent to which frictional force is capable of holding the bead.
The frictional force here provides the necessary centripetal force. If instantaneous angular velocity is ω then
By applying
We get ω =αt
A cubical block of side L rests on a rough horizontal surface with coefficient of friction μ . A horizontal force F is applied on the block as shown. If the coefficient of friction is sufficiently high so that the block does not slide before toppling, the minimum force required to topple the block is
The applied force shifts the normal reaction to one corner as shown. At this situation, the cubical block starts topping about O. Taking torque about O
A thin wire of length L and uniform linear mass density r is bent into a circular loop with centre at O as shown. The moment of inertia of the loop about the axis XX' is
Moment of inertia about the diameter of the circular loop (ring) = 1/2 MR^{2}
Using parallel axis theorem
The moment of inertia of the loop about XX' axis is
Where M = mass of the loop and R = radius of the loop Here M =Lρ and
An equilateral triangle ABC formed from a uniform wire has two small identical beads initially located at A. The triangle is set rotating about the vertical axis AO. Then the beads are released from rest simultaneously and allowed to slide down, one along AB and the other along AC as shown. Neglecting frictional effects, the quantities that are conserved as the beads slide down, are
The M.I. about the axis of rotation is not constant as the perpendicular distance of the bead with the axis of rotation increases.
Also since no external torque is acting.
Since, I increases, ω decreases.
One quarter sector is cut from a uniform circular disc of radius R. This sector has mass M. It is made to rotate about a line perpendicular to its plane and passing through the center of the original disc. Its moment of inertia about the axis of rotation is
The mass distribution of this sector is same about the axis of rotation as that of the complete disc about the axis. Therefore the formula remains the same as that of the disc.
A cylinder rolls up an inclined plane, reaches some height, and then rolls down (without slipping throughout these motions). The directions of the frictional force acting on the cylinder are
Imagine the cylinder to be moving on a frictionless surface. In both the cases the acceleration of the centre of mass of the cylinder is g sin q. This is also the acceleration of the point of contact of the cylinder with the inclined surface. Also no torque (about the centre of cylinder) is acting on the cylinder since we assumed the surface to be frictionless and the forces acting on the cylinder is mg and N which pass through the centre of cylinder. Therefore the net movement of the point of contact in both the cases is in the downward direction as shown. Therefore the frictional force will act in the upward direction in both the cases.
Note : In general we find the acceleration of the point of contact due to translational and rotational motion and then find the net acceleration of the point of contact.
The frictional force acts in the opposite direction to that of net acceleration of point of contact.
A circular platform is free to rotate in a horizontal plane about a vertical axis passing through its centre. A tortoise is sitting at the edge of the platform. Now, the platform is given an angular velocity ω_{0}. When the tortoise move along a chord of the platform with a constant velocity (with respect to the platform), the angular velocity of the platformω(t) will vary with time t as
Since there is no external torque, angular momentum remains conserved. As moment of inertia initially decreases and then increases, so ω will increase initially and then decreases.
Note : The M.I. of the system decreases when the tortoise move from A to B and then increases from B to A.
So the variation of w is nonlinear.
Consider a body, shown in figure, consisting of two identical balls, each of mass M connected by a light rigid rod. If an impulse J=MV is imparted to the body at one of its ends, what would be its angular velocity?
Change in angular momentum of the system = angular impulse given to the system about the centre of mass (Angular momentum)_{f }– (Angular momentum)_{i}
Let the system starts rotating with the angular velocity ω.
Moment of Inertia of the system about its axis of rotation [centre of mass of the system]
A particle undergoes uniform circular motion. About which point on the plane of the circle, will the angular momentum of the particle remain conserved?
The net force acting on a particle undergoing uniform circular motion is centripetal force which always passes through the centre of the circle. The torque due to this force about the centre is zero, therefore, L is conserved
about O.
A horizontal circular plate is rotating about a vertical axis passing through its centre with an angular velocity ω_{o}. A man sitting at the centre having two blocks in his hands stretches out his hands so that the moment of inertia of the system doubles. If the kinetic energy of the system is K initially, its final kinetic energy will be
KEYCONCPET : (K.E.)_{rotation} =L^{2}/2I
Here , L = constant
∴ (K.E.)_{rotational} × I = constant.
When I is doubled, K.E._{rotational }becomes half.
A disc is rolling without slipping with angular velocity w. P and Q are two points equidistant from the centre C. The order of magnitude of velocity is
Note : In pure rolling, th e poin t of conta ct is t he instantaneous centre of rotation of all the particles of the disc. On applying v = rω
We find ω is same for all the particles then v α r.
Farther the particles from O, higher is its velocity.
A block of mass m is at rest under the action of force F against a wall as shown in figure. Which of the following statement is incorrect?
The cubical block is in equilibrium.
For translational equilibrium (a) ∑F_{x} = 0 ⇒ F = N
(b) ∑F_{y} = 0 ⇒ f = mg
For Rotational Equilibrium
Where τ_{c} = torque about c.m.
Torque created by frictional force (f) about C = f^{×} a in clockwise direction.
There should be another torque which should counter this torque. The normal reaction N on the block acts as shown. This will create a torque N × b in the anticlockwise direction.
Such that f × a = N × b Note : The normal force does not act through the centre of the body always. The point of application of normal force depends on all the forces acting on the body.
From a circular disc of radius R and mass 9M, a small disc of radius R/3 is removed from the disc. The m oment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is
Let σ be the mass per unit area
The total mass of the disc
=σ×πR^{2} = 9M
The mass of the circular disc cut
Let us consider the above system as a complete disc of mass 9M and a negative mass M super imposed on it.
Moment of inertia (I1) of the complete disc = 1/2 9MR^{2} about a n a xis pa ssi n g th r ough O and perpendicular to the plane of the disc.
M.I. of the cut out portion about an axis passing through O' and perpendicular to the plane of disc
∴ M.I. (I_{2}) of the cut out portion about an axis passing through O and perpendicular to the plane of disc
[Using perpendicular axis theorem] ∴ The total M.I. of the system about an axis passing through O and perpendicular to the plane of the disc is I = I_{1} + I_{2}
A particle is confined to rotate in a circular path decreasing linear speed, then which of the following is correct?
Since v is changing (decreasing), L is not conserved in magnitude. Since it is given that a particle is confined to rotate in a circular path, it cannot have spiral path.
Since the particle has two accelerations a_{c} and a_{t} therefore the net acceleration is not towards the centre.
The direction of L remains same even when the speed decreases.
A solid sphere of mass M and radius R having moment of inertia I about its diameter is recast into a solid disc of radius r and thickness t. The moment of inertia of the disc about an axis passing the edge and perpendicular to the plane remains I. Then R and r are related as
For solid sphere
For solid disc
.
From (i) and (ii),
A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of 3v^{2}/4g with respect to the initial position. The object is
By the concept of energy conservation
This is the formula of the moment of inertia of the disc.
A bob of mass M is suspended by a massless string of length L. The horizontal velocity v at position A is just sufficient to make it reach the point B. The angle θ at which the speed of the bob is half of that at A, satisfies –
This is the case of vertical motion when the body just completes the circle. Here v = √5gL ....(i)
Applying energy conservation, Total energy at A = Total energy at P
Therefore, the value of q lies in the range
Look at the drawing given in the figure which has been drawn with ink of uniform linethickness. The mass of ink used to draw each of the two inner circles, and each of the two line segments is m. The mass of the ink used to draw the outer circle is 6 m.The coordinates of the centres of the different parts are: outer circle (0, 0), left inner circle (– a, a), right inner circle (a, a), vertical line (0, 0) and horizontal line (0, – a). The ycoordinate of the centre of mass of the ink in this drawing is
The system is made up of five bodies (three circles and two straight lines) of uniform mass distribution. Therefore we assume the system to be made up of five point masses where the mass of each body is considered at its geometrical centre.
The ycoordinate of the centre of mass is
A small mass m is attached to a massless string whose other end is fixed at P as shown in the figure. The mass is undergoing circular motion in the xy plane with centre at O and constant angular speed w. If the angular momentum of the system, calculated about O and P are denoted by respectively, then
The angular momentum of the mass m about O is mr^{2}ω and is directed toward +z direction for all locations of m.
The angular momentum of mass m about P is mvl and is directed for the given location of m as shown in the figure.
The direction of remains changing for different locations of m.
A thin uniform rod, pivoted at O, is rotating in the horizontal plane with constant angular speed ω, as shown in the figure. At time t = 0, a small insect starts from O and moves with constant speed v, with respect to the rod towards the other end. It reaches the end of the rod at t = T and stops. The angular speed of the system remains w throughout. The magnitude of the torque about O, as a function of time is best represented by which plot?
We know that
From the situation it is clear that the moment of inertia for (rod + insect) system is increasing.
Let at any instant of time 't', the insect is at a distance x from O. At this instant, the moment of inertia of the system is
From (i) & (ii)
When the insect stops moving, ur does not change and therefore τ becomes constant.
A uniform wooden stick of mass 1.6 kg and length l rests in an inclined manner on a smooth, vertical wall of height h(< l) such that a small portion of the stick extends beyond the wall. The reaction force of the wall on the stick is perpendicular to the stick. The stick makes an angle of 30° with the wall and the bottom of the stick is on a rough floor.
The reaction of the wall on the stick is equal in magnitude to the reaction of the floor on the stick. The ratio h/l and the frictional force f at the bottom of the stick are (g = 10 m s^{–2})
Considering the normal reaction of the floor and wall to be N and with reference to the figure.
By vertical equilibrium.
By horizontal equilibrium
Taking torque about A we get
1.6g × AB = N × x
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