Question

Two blocks, one of mass M and one of mass 3M, are connected by a massless string over a pulley that is a uniform disk of mass 2M and moment of inertia MR^2. The two masses are released from rest, and the masses accelerate as the pulley rotates. Assume there is negligible friction between the pulley and the axle. What is the linear acceleration, a, of the masses?

Answer 1

Answer:

4.9 m/s²

Explanation:

Let T be the tension in the string

If a is the linear acceleration in the direction of the 3M mass, the equation of motion on the 3M mass is

3Mg - T = 3Ma  (1)

Since the mass M moves upwards, its equation of motion is

T - Mg = Ma      (2)

From (2)

T = Ma + Mg

substituting T into (1), we have

3Mg - (Ma + Mg) = 3Ma

3Mg - Ma - Mg = 3Ma

collecting like terms, we have

3Mg - Mg = 3Ma + Ma

2Mg = 4Ma

dividing both sides by 4M, we have

2Mg/4M = 4Ma/4M

g/2 = a

a = g/2

Since g = 9.8 m/s²,

a = 9.8 m/s²/2

a = 4.9 m/s²

Answer 2

The linear acceleration 'a' of the masses M and 3M is; a = 4.9 m/s²

We are told that;

Mass of first block = M

Mass of second block = 3M

Let the tension in the strings be T.

Now for first block we can write;

T - Mg = Ma - - - (1)

For second block, we can write;

3Mg - T = 3Ma - - - (2)

Where a is linear acceleration of the masses.

Let us add eq 1 to eq 2 to get;

T - Mg + 3Mg - T = Ma + 3Ma

2Mg = 4Ma

M will cancel out to get;

2g = 4a

Using division property of equality by dividing both sides by 2 to get;

g = 2a

Thus;

a = g/2

Where g is acceleration due to gravity = 9.8 m/s²

Thus;

a = 9.8/2

a = 4.9 m/s²

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