The mass per unit length of a non-uniform rod of length L is given by where is a constant and x is the distance from one end of the rod. The distance between the centre of mass of the rod and this end is:
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2.
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4.
At t = 0, the positions of the two blocks are shown. There is no external force acting on the system. Find the coordinates of the center of mass of the system at t = 3 seconds:
1. | (1, 0) | 2. | (3, 0) |
3. | (4.5, 0) | 4. | (2.25, 0) |
A uniform square plate ABCD has a mass of 10 kg.
If two point masses of 5 kg each are placed at the corners C and D as shown in the adjoining figure, then the centre of mass shifts to the mid-point of:
1. OH
2. DH
3. OG
4. OF
A rod is falling down with constant velocity \(V_0\) as shown. It makes contact with hinge A and rotates around it. The angular velocity of the rod just after the moment when it comes in contact with hinge A is:
1. | \(2 \mathrm{V}_0 / 3 \mathrm{L} \) | 2. | \(3 \mathrm{V}_0 / 2 \mathrm{L} \) |
3. | \(\mathrm{V}_0 / \mathrm{L} \) | 4. | \(2 \mathrm{V}_0 / 5 \mathrm{L}\) |
The law of conservation of angular momentum is valid when:
1. | The net force is zero and the net torque is non-zero | 2. | The net force is non-zero and the net torque is non zero |
3. | Net force may or may not be zero and net torque is zero | 4. | Both force and torque must be zero |
Particles A and B are separated by 10 m, as shown in the figure. If A is at rest and B started moving with a speed of 20 m/s then the angular velocity of B with respect to A at that instant is:
1. | 1 rad s-1 | 2. | 1.5 rad s-1 |
3. | 2 rad s-1 | 4. | 2.5 rad s-1 |
The value of M, as shown, for which the rod will be in equilibrium is:
1. | 1 kg | 2. | 2 kg |
3. | 4 kg | 4. | 6 kg |
In the three figures, each wire has a mass M, radius R and a uniform mass distribution. If they form part of a circle of radius R, then about an axis perpendicular to the plane and passing through the centre (shown by crosses), their moment of inertia is in the order:
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2.
3.
4.
1. | \(\vec{\tau}=(-17 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+4 \widehat{\mathrm{k}})\) N-m |
2. | \(\vec{\tau}=(-17 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-4 \widehat{\mathrm{k}}) \) N-m |
3. | \(\vec{\tau}=(17 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+4 \widehat{\mathrm{k}})\) N-m |
4. | \(\vec{\tau}=(-41 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+16 \hat{\mathrm{k}})\) N-m |
Four thin rods, each of mass m and length L, form a square. The moment of inertia on any side of the square is:
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