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Since we know that :                                             Differentiating the equation with respect to time, we obtain :                                             or                                                   or                                                                                                                 Now using Newton's law of motion we can write :         ...
The position vector of the ith particle (with respect to the center of mass) is given by :                                                                      or                                                                   From the first case we can write :                                                                     Taking cross product with position vector we get  ;              ...
From the first part we can write :                                                          or                                                      Squaring both sides (in vector form taking dot products with itself), we get :                                                           or                                                      or                                                     ...
The momentum of ith particle is given by :              The velocity of the centre of mass is V. Then the velocity of ith particle with respect to the center of mass will be :      Now multiply the mass of the particle to both the sides, we get :                                                                 or                                                                                    ...
True. Since it is a frictionless plane so frictional force is zero thus torque is not generated. This results in slipping not rolling.
True. As the frictional force at the bottommost point is zero, so the work done against it is also zero.
False. The value of acceleration at contact has some value, it is not zero as the frictional force is zero but the force applied will give some acceleration.
True. This is because the translational speed is balanced by rotational speed.
False. Friction also opposes the relative motion between the contacted surfaces. In the case of rolling, the cm is moving in backward direction thus the frictional force is directed in the forward direction.
In case of rolling without any skidding is given by :                                               Thus                                   or                                                       or
We know that the bottommost point of body (which is in contact with the surface) is at rest during rolling. Thus work done against the frictional force is zero.
Consider the following figure :                                        Moment of inertia of the cylinder is   :                                                                Thus acceleration is given by :                                                                or                                                                  Now using Newton's law of motion :                        ...
Friction is the cause for motion here. So using Newton's law of motion we can write :                                                                                        or                                            Now by the equation of motion, we can write :                                                or                                             The torque is given by :              ...
Perfect rolling will occur when the velocity of the bottom point (B) will be zero. Thus the frictional force acting will be zero.
Since the velocity at point B is tangentially leftward so the frictional force will act in the rightward direction. The sense of frictional torque is perpendicular (outward) to the plane of the disc.
Let the angular speed of the disc is  . So the linear velocity can be written as     (a)  Point A:-                                  The magnitude of linear velocity is  and it is tangentially rightward. (b)  Point B:-                                   The magnitude linear velocity is  and its direction is tangentially leftward.  (c) Point C:-                                ...
Consider the given situation :                                                The total energy when the object is at the top (potential energy)  =  mgh. Energy when the object is at the bottom of the plane :                                                                                Put           and      , we get :                                                                             ...
Consider the figure given below :                                                   The moment of inertia about RS axis:-                                       Now the moment of inertia about QP axis:-                                  or                                         or                                         Thus                                                                        ...
Consider the figure given below:                                           The moment of inertia about the x-axis is given by :                                                     And the moment of inertia about the y-axis is :                                                     Now about z-axis :                                           or
Let the moment of inertia of disc I and disc II be I1 and I2 respectively. Similarly, the angular speed of disc I and disc II be w1 and w2 respectively. So the angular momentum can be written as :                                                        and              Thus the total initial angular momentum is :       Now when the two discs are combined the angular momentum is :                ...
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