A system consists of two blocks connected by a light string that passes over a massive, uniform disk pulley. Block A (mass mA) rests on a frictionless horizontal table. Block B (mass mB) hangs vertically, connected to block A by the string. The pulley has mass Mp and radius R, and it rotates without friction about its axle. The string does not slip on the pulley. What is the magnitude of the acceleration of block A?
Correct: C
Let 'a' be the magnitude of the acceleration of both blocks. Since the string does not slip and is inextensible, the angular acceleration of the pulley is α = a/R. Let T1 be the tension in the string between block A and the pulley, and T2 be the tension in the string between block B and the pulley. Since the pulley has mass, T1 and T2 will not be equal.
1. **For Block A (horizontal motion):**
Applying Newton's Second Law: ΣFx = mA * a
T1 = mA * a (Equation 1)
2. **For Block B (vertical motion):**
Applying Newton's Second Law: ΣFy = mB * a
mB * g - T2 = mB * a (Equation 2)
3. **For the Pulley (rotational motion):**
The pulley is a uniform disk, so its moment of inertia is I = (1/2)Mp * R².
The net torque on the pulley is τ_net = (T2 - T1)R. (T2 creates a torque in the direction of motion, T1 opposes it).
Applying Newton's Second Law for rotation: τ_net = I * α
(T2 - T1)R = (1/2)Mp * R² * (a/R)
(T2 - T1)R = (1/2)Mp * R * a
Dividing by R (assuming R ≠ 0):
T2 - T1 = (1/2)Mp * a (Equation 3)
Now, we have a system of three equations with three unknowns (T1, T2, a). We want to solve for 'a'.
From Equation 1, substitute T1 into Equation 3:
T2 - (mA * a) = (1/2)Mp * a
T2 = (mA * a) + (1/2)Mp * a
T2 = (mA + (1/2)Mp) * a (Equation 4)
Now, substitute T2 from Equation 4 into Equation 2:
mB * g - (mA + (1/2)Mp) * a = mB * a
mB * g = mB * a + (mA + (1/2)Mp) * a
mB * g = (mB + mA + (1/2)Mp) * a
mB * g = (mA + mB + (1/2)Mp) * a
Solving for 'a':
a = (mB * g) / (mA + mB + (1/2)Mp)
The correct choice is C.