A block of mass m = 9.7 kg hangs from the end of a massless rope wound around a 30.3 kg solid uniform cylinder which has a radius of 0.7 m. The cylinder can rotate freely about a frictionless axle through its center. Assuming the rope does not slip on the cylinder, what is the magnitude of the acceleration of m, in units of m/s2 ? Enter the numerical value only, do not enter units. Enter your answer to the nearest one decimal place, i.e., X.X

Question

A block of mass m = 9.7 kg hangs from the end of a massless rope wound around a 30.3 kg solid uniform cylinder which has a radius of 0.7 m. The cylinder can rotate freely about a frictionless axle through its center.  Assuming the rope does not slip on the cylinder, what is the magnitude of the acceleration of m, in units of m/s2 ? Enter the numerical value only, do not enter units. Enter your answer to the nearest one decimal place, i.e., X.X

BlackTom AI · Accepted Answer

The system consists of a hanging mass m = 9.7 kg connected by a rope that does not slip on a solid uniform cylinder of mass M = 30.3 kg and radius R = 0.7 m. The rope is massless and the axle is frictionless, so the cylinder can rotate as the mass moves.

First, relate the linear acceleration a of the mass to the angular acceleration of the cylinder. Since the rope does not slip, a = R * alpha, where alpha is the angular acceleration of the cylinder.

For the mass m: the forces are gravity downward (mg) and tension T upward. Taking downward as positive, mg - T = m a.  (Equation 1)

For the cylinder: the torque due to the tension T causes angular acceleration. The torque is T * R, and it equals I * alpha, where I is the moment of inertia of a solid uniform cylinder, I = (1/2) M R^2. Substituting alpha = a / R gives T * R = I * (a / R) = (1/2) M R^2 * (a / R) = (1/2) M R a.

From this, solve for T: T = [(1/2) M R a] / R = (1/2) M a. So the tension is proportional to the cylinder’s mass and the linear acceleration, but independent of the radius in this arrangement.

Now substitute T into Equation (1): mg - (1/2) M a = m a. Rewrite as mg = a (m + M/2). Therefore, a = mg / (m + M/2).

Plug in the given values: m = 9.7 kg, M = 30.3 kg, g = 9.8 m/s^2.
- M/2 = 15.15 kg
- Denominator: m + M/2 = 9.7 + 15.15 = 24.85
- Numerator: m g = 9.7 × 9.8 = 95.06
- Acceleration a = 95.06 / 24.85 ≈ 3.825 m/s^2

Rounding to the nearest one decimal place gives approximately 3.8 m/s^2.  (Note: if a student incorrectly treats the cylinder as a point mass or neglects its rotational inertia, they might obtain a larger value such as around 4.7 m/s^2, which misaccounts for the rotational inertia of the cylinder.)

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