Advertisements
Advertisements
प्रश्न
A flywheel of moment of inertia 5⋅0 kg-m2 is rotated at a speed of 60 rad/s. Because of the friction at the axle it comes to rest in 5⋅0 minutes. Find (a) the average torque of the friction (b) the total work done by the friction and (c) the angular momentum of the wheel 1 minute before it stops rotating.
Advertisements
उत्तर
Let the angular deceleration produced due to frictional force be α.
Initial angular acceleration,
\[\omega_0 = 60 rad/s\]
Final angular velocity,
\[\omega = 0\]
t = 5 min =300 s
We know that
\[\omega = \omega_0 + \alpha t\]
\[\Rightarrow \alpha = - \frac{\omega_0}{t}\]
\[ \Rightarrow \alpha = - \left( \frac{60}{300} \right) = - \frac{1}{5} rad/ s^2\]
(a) Torque produced by the frictional force (R),
\[\tau = I\alpha = 5 \times $\left( \frac{- 1}{5}
right)\]
= 1 N - m opposite to the rotation of wheel
(b) By conservation of energy,
Total work done in stopping the wheel by frictional force = Change in energy
\[W = \frac{1}{2}I \omega^2 \]
\[ = \frac{1}{2} \times 5 \times \left( 60 \times 60 \right)\]
\[ = 9000 \text{ joule }= 9 kJ\]
(c) Angular velocity after 4 minutes,
\[\omega = \omega_0 + \alpha t\]
\[ = 60 - \frac{4 \times 60}{5}\]
\[ = \frac{60}{5} = 12 rad/s\]
So, angular momentum about the centre,
\[L = I\omega\]
\[ = 5 \times 12 = 60 kg - m^2 /s\]
APPEARS IN
संबंधित प्रश्न
Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components px, py and 'p_z`. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.
A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s–1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of the angular momentum of the cylinder about its axis?
If several forces act on a particle, the total torque on the particle may be obtained by first finding the resultant force and then taking torque of this resultant. Prove this. Is this result valid for the forces acting on different particles of a body in such a way that their lines of action intersect at a common point?
If the resultant torque of all the forces acting on a body is zero about a point, is it necessary that it will be zero about any other point?
A ladder is resting with one end on a vertical wall and the other end on a horizontal floor. If it more likely to slip when a man stands near the bottom or near the top?
When a body is weighed on an ordinary balance we demand that the arum should be horizontal if the weights on the two pans are equal. Suppose equal weights are put on the two pans, the arm is kept at an angle with the horizontal and released. Is the torque of the two weights about the middle point (point of support) zero? Is the total torque zero? If so, why does the arm rotate and finally become horizontal?
A simple pendulum of length l is pulled aside to make an angle θ with the vertical. Find the magnitude of the torque of the weight ω of the bob about the point of suspension. When is the torque zero?
A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N?
State conservation of angular momentum.
Choose the correct alternatives:
- For a general rotational motion, angular momentum L and angular velocity ω need not be parallel.
- For a rotational motion about a fixed axis, angular momentum L and angular velocity ω are always parallel.
- For a general translational motion , momentum p and velocity v are always parallel.
- For a general translational motion, acceleration a and velocity v are always parallel.
Figure shows two identical particles 1 and 2, each of mass m, moving in opposite directions with same speed v along parallel lines. At a particular instant, r1 and r2 are their respective position vectors drawn from point A which is in the plane of the parallel lines. Choose the correct options:
- Angular momentum l1 of particle 1 about A is l1 = mvd1
- Angular momentum l2 of particle 2 about A is l2 = mvr2
- Total angular momentum of the system about A is l = mv(r1 + r2)
- Total angular momentum of the system about A is l = mv (d2 − d1)
⊗ represents a unit vector coming out of the page.
⊗ represents a unit vector going into the page.
A rod of mass 'm' hinged at one end is free to rotate in a horizontal plane. A small bullet of mass m/4 travelling with speed 'u' hits the rod and attaches to it at its centre. Find the angular speed of rotation of rod just after the bullet hits the rod 3. [take length of the rod as 'l']
The position vector of 1 kg object is `vecr = (3hati - hatj)` m and its velocity `vecv = (3hati + hatk)` ms−1. The magnitude of its angular momentum is `sqrtx` Nm where x is ______.
A particle of mass ‘m’ is moving in time ‘t’ on a trajectory given by
`vecr = 10alphat^2hati + 5beta(t - 5)hatj`
Where α and β are dimensional constants.
The angular momentum of the particle becomes the same as it was for t = 0 at time t = ______ seconds.
Angular momentum of a single particle moving with constant speed along the circular path ______.
