Motion of rigid body - class-XI

For a sphere on a smooth surface, the acceleration of the center of mass is given by a = F/m, according to Newton's second law. This value is independent of the point of application 'h'.

Let the mass distribution on the left of seesaw be x kg and that on the right be y kg such that $x>y$. And let the mass of the girl be $m _g$ and that of heavy object be $m _h$ such that $m _h>m _g$
Now during the equilibrium position we have the moments about the fulcrum as
$m _h\times \displaystyle\frac{l}{2}+y\times \displaystyle\frac{l}{4}=m _g\times \displaystyle\frac{l}{2}+x\times \displaystyle\frac{l}{4}$
or
$m _h-m _g=\displaystyle\frac{x-y}{2}$
Now when both the girl and the heavy object move half the distance we see that the moment about the pivot by the girl and the heavy object are $m _g\times \displaystyle\frac{l}{4}+x\times \displaystyle\frac{l}{4}$ and $m _h\times \displaystyle\frac{l}{4}+y\times \displaystyle\frac{l}{4}$

Subtracting moment of the girl from the moment of the heavy object
$(m _h+y)\displaystyle\frac{l}{4}-(m _g+x)\displaystyle\frac{l}{4}$
or
$\displaystyle\frac{l}{4}(m _h-m _g+y-x)$
or
$\displaystyle\frac{l}{4}(\displaystyle\frac{x}{2}-\displaystyle\frac{y}{2}+y-x)$
or
$\displaystyle\frac{l}{4}(\displaystyle\frac{y-x}{2})$
We get this value as negative because $x>y$
Thus the moment applied by the girl is more than that of the heavy object implying that the side of the girl tilts downwards

Since all the weights fall freely under gravity along with the assembly of the rod, the acceleration of all objects is equal to $g$. Hence all the weights have same relative distance between them at all points, and hence no stretching of spring occurs and thus the position(distance of falling) of all the weights would be the same.

A wheel is rotated with uniform angular velocity even if the tangential force is applied it happens only if a counter torque is acted on the wheel to balance the torque of the applied force. The $ angular$ $ acceleration $ of the wheel is $ zero$ if the torque of the tangential force is balanced by any other force $(i.e. friction)$.

To keep the rod horizontal, total torque should be zero

Since, The energy of cone resting on its side is minimum, it will be in neutral equilibrium.

Since curved and heavy base provide stability to the toys so they have curved and heavy bases.

In order to lift the load we have to apply force such that it can balance the torque by load.

The relationship v = omega * r implies that for a constant linear speed v, omega is inversely proportional to r (omega = v/r).

When the top slows down, air bouancy, gravitational force, inertial movement and other complicated motion concepts act on it resulting into wobbling of the top.

Net external force acting on a pure rotating body is zero i.e.  $F _{ext} = 0$

Initially the origin was (0,0) and the point was (3,4). If the origin is now shifted to a new point (1,3), then the coordinates of the new point will be (X,Y)  with respect to the new origin(h=1, k=3).  Thus, (3,4) = (X+1, Y+3). Solving, we get, X=2 and Y=1

A bicycle tyre rotates around the center which causes forward motion of the bicycle. Therefore, both rotation and linear motion of tyre is involved. 

In circular motion, the body revolves around a certain axis. So that body posses a certain angular velocity, angular acceleration and frequency of rotation where these terms are used in rotational motion. Thus circular motion is said to be similar to rotational motion.

Spinning of top, spinning wheel and spinning of the earth are purely rotational motion whereas rolling ball exhibits both rotational motion as well as translational motion. As precession is described by rotational motion only, thus rolling of ball cannot be considered as an example of precession.

The force is F = -5(x-2)^2. For motion to be oscillatory, there must be a stable equilibrium point and a restoring force. Since the force is proportional to the square of the displacement, it is not SHM (which requires F proportional to x), but it is oscillatory as it acts to restore the particle toward the equilibrium position x=2.

The force F = -5(x-2) is a linear restoring force of the form F = -k(x-x0). This is the definition of Simple Harmonic Motion (SHM). Since all SHM is also oscillatory, both options (2) and (3) are correct.

A single external force acting on a body will cause both translational and rotational acceleration if the line of action does not pass through the center of mass. A body can be in pure rotation only if the net force is zero or if the body is constrained (like a hinge), so Ram's statement is incorrect. Shyam's statement is also incorrect because pure rotation is not exclusive to rigid bodies, and a rigid body can still undergo translation under a single force.

Torque $\left( \tau  \right) =I\alpha $
$\tau =\cfrac { 1 }{ 2 } \times M{ R }^{ 2 }\times \cfrac { 2\pi \left( { n } _{ 2 }-{ n } _{ 1 } \right)  }{ t } $
$\therefore$ $\tau =16\times { \left( \cfrac { 1 }{ 2 }  \right)  }^{ 2 }\times \pi \cfrac { \left( 2-0 \right)  }{ 8 } =\pi N-m$

Basic theory ,$(dx,dy) $ is called as transitional or shift vector.

Rotational motion and translational motion both are constrained motion.

Any point on the rigid body, undergoing rolling without slipping, will have linear and angular speed, such that

A spinning top is a common example of precession.

When a body is rolling without slipping on the ground, its center of mass exhibits translational motion whereas the body exhibits rotational motion in its center of mass frame. Thus rolling body exhibits both translational motion as well as rotational motion.

 Additional force is Coriolis force which acts perpendicular to the plane of motion. so $\Delta T=2mv\omega $ .