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Two particles A and B initially at rest, move towards each other under mutual force of attraction. At the instant when the speed of A is V and the speed of B is 2V, the speed of the centre of mass of the system is
We know that Fext = Mac.m. ... (i)
We consider the two particles in a system. Mutual force of attraction is a internal force. There are no external forces acting on the system. From (i)
ac.m. = 0 ⇒ vc.m. = constant.
Since, initially the vc.m. = 0
∴ Finally vc.m. = 0
A mass M moving with a constant velocity parallel to the X-axis. Its angular momentum with respect to the origin
L = Momentum × perpendicular distance of line of action of momentum w.r.t point of rotation
L = Mv × y.
The quantities on the right side of the equation are not changing.
The magnitude is constant. The direction is also constant.
When a bicycle is in motion, the force of friction exerted by the ground on the two wheels is such that it acts
When the cycle is not pedalled but the cycle is in motion (due to previous effort) the wheels move in the direction such that the centre of mass of the wheel move forward. Rolling friction will act in the opposite direction to the relative motion of the centre of mass of the body with respect to ground. Therefore the rolling friction will act in backward direction in both the wheels. The sliding friction will act in the forward direction of rear wheel during pedalling.
A particle of mass m is projected with a velocity v making an angle of 45° with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height h is
Angular momentum = (momentum) × (perpendicular distance of the line of action of momentum from the axis of rotation) Angular momentum about O
From (i) and (ii)
Also, from (i) and (ii)
A uniform bar of length 6a and mass 8m lies on a smooth horizontal table. Two point masses m and 2m moving in the same horizontal plane with speed 2v and v, respectively, strike the bar [as shown in the fig.] and stick to the bar after collision. Denoting angular velocity (about the centre of mass), total energy and centre of mass velocity by w, E and vc respectively, we have after collision
Applying conservation of linear momentum 2m (– v) + m (2v) + 8m × 0 = (2m + m + 8m) vc
⇒ vc = 0
Applying conservation of angular momentum about centre of mass
I = 30ma2 ...(ii)
From (i) and (ii)
Energy after collision,
The moment of inertia of a thin square plate ABCD, Fig., of uniform thickness about an axis passing through the centre O and perpendicular to the plane of the plate is
wh ere I1, I 2 , I3 andI4 are respectively the momen ts of intertial about axis 1, 2, 3 and 4 which are in the plane of the plate.
To find the moment of inertia of ABCD about an axis passing through the centre O and perpendicular to the plane of the plate, we use perpendicular axis theorem. If we consider ABCD to be in the X-Y plane then we know that
Also, Izz' = I3 + I4
Adding (i) and (ii),
But I1 = I2 and I3 = I4
A tube of length L is filled completely with an incomressible liquid of mass M and closed at both the ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity ω. The force exerted by the liquid at the other end is
The force acting on the mass of liquid dm of length dx at a distance x from the axis of rotation O.
dF = (dm) x ω2
where M/L is mass of liquid in unit length.
∴The force acting at the other end is for the whole liquid in tube.
A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 m/s. A pendulum bob is suspended from the roof of the car by a light rigid rod of length 1.00 m. The angle made by the rod with track is
When the car is moving in a circular horizontal track of radius 10 m with a constant speed, then the bob is also undergoing a circular motion. The bob is under the influence of two forces.
(i) T (tension in the rod)
(ii) mg (weight of the bob)
Resolving tension, we get
Tcosθ = mg ... (i)
And Tsinθ = mv2/r ... (ii)
(Here T sin θ is producing the necessary centripetal force for circular motion)
Dividing (ii) by (i), we get
Let I be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle q with AB. The moment of inertia of the plate about the axis CD is then equal to
A'B' ⊥ AB and C' D'⊥ CD
From symmetry IAB = IA'B' and ICD = IC'D'
From theorem of
Izz = IAB + IA'B' = ICD + IC'D'
⇒ 2IAB = 2ICD
∴ IAB = ICD
The torqueτ on a body about a given point is found to be equal to A × L where A is a constant vector, and L is the angular momentum of the body about that point. From this it follows that
From cross-product rule, uur is always perpendicular to the
By the dot product definition
Differentiating with respect to time
⇒ L = constant
Thus, the magnitude of L always remains constant.
A solid cylinder is rolling down a rough inclined plane of inclination θ. Then
As shown in the figure, the component of weight mg sinθ tends to slide the point of contact (of the cylinder with inclined plane) along its direction. The sliding friction acts in the opposite direction to oppose this relative motion.
Because of frictional force the cylinder rolls.
Thus frictional force adds rotation but hinders translational motion.
Applying Fnet = ma along the direction of inclined plane,
we get mg sin
where ac = acceleration of centre of mass of the cylinder
From (i) and (ii)
If θ is reduced, frictional force is reduced.
If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that
Due to internal forces acting in the system, the kinetic and potential energy may change with time.
Also zero external force may create a torque if the line of action of forces are along different direction. Thus the torque will change the angular momentum of the system.
A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, A is the point of contact, B is the centre of the sphere and C is its topmost point. Then,
is the velocity of centre of the sphere, then
(b) is the correct option.
(c) is the correct option.
Two solid spheres A and B of equal volumes but of different densities dAand dBare connected by a string. They are fully immersed in a fluid of density dF. They get arranged into an equilibrium state as shown in the figure with a tension in the string. The arrangement is possible only if
Let V be the volume of spheres.
For equilibrium of A :
T + VdAg = Vdfg
(a) is the correct option
For equilibrium of B :
(b) is the correct option
(d) is the correct option.
A thin ring of mass 2 kg and radius 0.5 m is rolling without on a horizontal plane with velocity 1 m/s. A small ball of mass 0.1 kg, moving with velocity 20 m/s in the opposite direction hits the ring at a height of 0.75 m and goes vertically up with velocity 10 m/ s. Immediately after the collision
The frictional force between
the ring and the ball is impulsive. The angular impulse created by this force tends to decrease the angular speed of the ring about O.
After the collision the angular speed decreases but the ring remains rotating in the anticlock wise direction.
Therefore the friction between the ring and the ground (at the point of contact) is towards left.
The figure shows a system consisting of (i) a ring of outer radius 3R rolling clockwise without slipping on a horizontal surface with angular speed ω and (ii) an inner disc of radius 2R rotating anti-clockwise with angular speed ω/2. The ring and disc are separated by frictionless ball bearings. The point P on the inner disc is at a distance R from the origin, where OP makes an angle of 30° with the horizontal. Then with respect to the horizontal surface,
For rolling motion, the velocity of the point of contact with respect to the surface should be zero. For this
A shown in the figure, the point P will have two velocities
making an angle of 30º with the vertical due to rotation
Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most of its mass concentrated near the axis. Which statement(s) is(are) correct?
The acceleration of the center of mass of cylinder rolling down an inclined plane is
Here IP > IQ because in case of P the mass is concentrated away from the axis.
In the figure, a ladder of mass m is shown leaning against a wall. It is in static equilibrium making an angleθ with the horizontal floor. The coefficient of friction between the wall and the ladder is μ1 and that between the floor and the ladder is μ2. The normal reaction of the wall on the ladder is N1 and that of the floor is N2. If the ladder is about to slip, then
Solving the above equation we get
∴ (c) is the correct option.
When μ1 = 0
Taking torque about P we get
∴ (d) is correct
A ring of mass M and radius R is rotating with angular speed ω about a fixed vertical axis passing through its centre O with two point masses each of mass M/8 at rest at O. Thesemasses can move radially outwards along two massless rods fixed on the ring as shown in the figure. At some instant the angular speed of the system is 8/9 ω and one of the masses is
at a distance of 3/5 R from O. At this instant the distance of the other mass from O is
Applying conservation of angular mumentum about the axis
D is the correct option
Two thin circular discs of mass m and 4m, having radii of a and 2a, respectively, are rigidly fixed by a massless, rigid rod of length through their centres. This assembly is laid on a firm and flat surface, and set rolling without slipping on the surface so that the angular speed about the axis of the rod is ω. The angular momentum of the entire assembly about the point (see the figure)Which of the following statement(s) is (are) true?
The circumference of a circle of radius OM will be 2π(5a) = 10πa.
For completing this circle once, the smaller disc will have to take
Therefore the C.M. of the assembly rotates about zaxis with an angular speed of w/5.
The angular momentum about the C.M. of the system
The position of a particle of mass m is given by the following equation
where which of the following statement(s) is(are) true about the particle?
At t = 1s
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