The question is asking to match the given acceleration and angular velocity vectors (List-I) with the corresponding trajectories of the instantaneous center of rotation of the disc (List-II).

The instantaneous center of rotation of the disc can be found by using the formula for the Coriolis effect, which is given by 2m(vec{v} x vec{ω}), where m is the mass of the disc, vec{v} is the velocity of the disc, and vec{ω} is the angular velocity of the disc.

Let's solve this step by step:

1. For (I), vec{a} = -2j m/s² and vec{ω} = 4k rad/s. The acceleration is in the negative y-direction and the angular velocity is in the positive z-direction. This means the disc is moving downwards and rotating counterclockwise when viewed from above. The trajectory of the instantaneous center of rotation would be a straight line along the y-axis, so the correct match is (P) y = 4.

2. For (II), vec{a} = -2j m/s² and vec{ω} = -4k rad/s. The acceleration is in the negative y-direction and the angular velocity is in the negative z-direction. This means the disc is moving downwards and rotating clockwise when viewed from above. The trajectory of the instantaneous center of rotation would be a parabola opening to the left, so the correct match is (Q) y = 4x² (x ≤ 0).

3. For (III), vec{a} = -2i m/s² and vec{ω} = 4k rad/s. The acceleration is in the negative x-direction and the angular velocity is in the positive z-direction. This means the disc is moving to the left and rotating counterclockwise when viewed from above. The trajectory of the instantaneous center of rotation would be a parabola opening to the right, so the correct match is (R) y = 4x² (x ≥ 0).

4. For (IV), vec{a} = -2i m/s² and vec{ω} = -4k rad/s. The acceleration is in the negative x-direction and the angular velocity is in the negative z-direction. This means the disc is moving to the left and rotating clockwise when viewed from above. The trajectory of the instantaneous center of rotation would be a parabola opening upwards, so the correct match is (S) x = 4y² (y ≥ 0).

So, the correct option is (A) I → P, II → Q, III → R, IV → S.

Question

The question is asking to match the given acceleration and angular velocity vectors (List-I) with the corresponding trajectories of the instantaneous center of rotation of the disc (List-II).

The instantaneous center of rotation of the disc can be found by using the formula for the Coriolis effect, which is given by 2m(vec{v} x vec{ω}), where m is the mass of the disc, vec{v} is the velocity of the disc, and vec{ω} is the angular velocity of the disc.

Let's solve this step by step:

1. For (I), vec{a} = -2j m/s² and vec{ω} = 4k rad/s. The acceleration is in the negative y-direction and the angular velocity is in the positive z-direction. This means the disc is moving downwards and rotating counterclockwise when viewed from above. The trajectory of the instantaneous center of rotation would be a straight line along the y-axis, so the correct match is (P) y = 4.

2. For (II), vec{a} = -2j m/s² and vec{ω} = -4k rad/s. The acceleration is in the negative y-direction and the angular velocity is in the negative z-direction. This means the disc is moving downwards and rotating clockwise when viewed from above. The trajectory of the instantaneous center of rotation would be a parabola opening to the left, so the correct match is (Q) y = 4x² (x ≤ 0).

3. For (III), vec{a} = -2i m/s² and vec{ω} = 4k rad/s. The acceleration is in the negative x-direction and the angular velocity is in the positive z-direction. This means the disc is moving to the left and rotating counterclockwise when viewed from above. The trajectory of the instantaneous center of rotation would be a parabola opening to the right, so the correct match is (R) y = 4x² (x ≥ 0).

4. For (IV), vec{a} = -2i m/s² and vec{ω} = -4k rad/s. The acceleration is in the negative x-direction and the angular velocity is in the negative z-direction. This means the disc is moving to the left and rotating clockwise when viewed from above. The trajectory of the instantaneous center of rotation would be a parabola opening upwards, so the correct match is (S) x = 4y² (y ≥ 0).

So, the correct option is (A) I → P, II → Q, III → R, IV → S.

Knowee AI · Accepted Answer

The question is asking to match the given acceleration and angular velocity vectors (List-I) with the corresponding trajectories of the instantaneous center of rotation of the disc (List-II).

The instantaneous center of rotation of the disc can be found by using the formula for the Coriolis effect, which is given by 2m(vec{v} x vec{ω}), where m is the mass of the disc, vec{v} is the velocity of the disc, and vec{ω} is the angular velocity of the disc.

Let's solve this step by step:

1. For (I), vec{a} = -2j m/s² and vec{ω} = 4k rad/s. The acceleration is in the negative y-direction and the angular velocity is in the positive z-direction. This means the disc is moving downwards and rotating counterclockwise when viewed from above. The trajectory of the instantaneous center of rotation would be a straight line along the y-axis, so the correct match is (P) y = 4.

2. For (II), vec{a} = -2j m/s² and vec{ω} = -4k rad/s. The acceleration is in the negative y-direction and the angular velocity is in the negative z-direction. This means the disc is moving downwards and rotating clockwise when viewed from above. The trajectory of the instantaneous center of rotation would be a parabola opening to the left, so the correct match is (Q) y = 4x² (x ≤ 0).

3. For (III), vec{a} = -2i m/s² and vec{ω} = 4k rad/s. The acceleration is in the negative x-direction and the angular velocity is in the positive z-direction. This means the disc is moving to the left and rotating counterclockwise when viewed from above. The trajectory of the instantaneous center of rotation would be a parabola opening to the right, so the correct match is (R) y = 4x² (x ≥ 0).

4. For (IV), vec{a} = -2i m/s² and vec{ω} = -4k rad/s. The acceleration is in the negative x-direction and the angular velocity is in the negative z-direction. This means the disc is moving to the left and rotating clockwise when viewed from above. The trajectory of the instantaneous center of rotation would be a parabola opening upwards, so the correct match is (S) x = 4y² (y ≥ 0).

So, the correct option is (A) I → P, II → Q, III → R, IV → S.

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