Answer:
The answer is B. 4.71 m/s²
Step by step explanation:
To find the magnitude of the acceleration of the falling cylinder, we can consider the forces acting on it.
As the cylinder falls, two forces are acting on it: the gravitational force (mg) pulling it downward and the tension force (T) in the string pulling it upward.
Since the cylinder is massless, we can neglect its mass in calculations.
The tension force (T) in the string is related to the angular acceleration (α) of the cylinder as it unwinds. The tension force can be expressed as T = Iα, where I is the moment of inertia of the cylinder.
For a solid cylinder rotating about its axis, the moment of inertia is given by I = 0.5mr², where m is the mass of the cylinder and r is its radius.
The gravitational force (mg) is equal to the weight of the cylinder, which is given by mg = 0.400 kg * 9.8 m/s² = 3.92 N.
The tension force (T) and gravitational force (mg) are in opposite directions, so we can write the net force equation as T - mg = m * a, where a is the acceleration of the falling cylinder.
Substituting the expressions for T and mg, we have 0.5mr²α - mg = m * a.
Since α = a/r (for a solid cylinder rolling without slipping), we can rewrite the equation as 0.5maα - mg = m * a.
Simplifying the equation, we have 0.5aα - g = a.
Rearranging the equation, we get a * (0.5α - 1) = g.
Finally, solving for a, we have a = g / (0.5α - 1).
Given the radius r = 0.100 m, we can calculate α using α = a/r.
Let's substitute the values and calculate the acceleration:
α = a/r = (9.8 m/s²) / (0.5 * (9.8 m/s² / (0.100 m))) = 19.6 rad/s².
Substituting α into the equation for a, we have a = (9.8 m/s²) / (0.5 * 19.6 rad/s² - 1) ≈ 4.71 m/s².
Therefore, the magnitude of the acceleration of the falling cylinder is approximately 4.71 m/s².
The correct answer is B. 4.71 m/s².