A thin, light horizontal string is wrapped around the rim of a 4.00-kg solid uniform disk that is 30.0 cm in diameter. A box is connected to the right end of the string and moves to the right along the ground horizontally with no friction. The box is subjected to a horizontal force of magnitude F =100. 0 N parallel to the ground. The box has mass m = 1.00 kg. The disk is rotates clockwise about a fixed axis attached to a steel structure bolted to the ground:
(a) (20 points) What is a, the linear acceleration magnitude the box?
(b) (15 points) What is the tension T in the string?
(c) (5 points) If the disk is replaced with a hollow thin walled cylinder of the same mass, and diameter, what will be the acceleration in part (a)?
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To solve this problem, we can use the principles of Newton's laws of motion and rotational motion.
(a) To find the linear acceleration magnitude of the box, we can use Newton's second law, which states that the net force on an object is equal to the product of its mass and acceleration (F = m * a).
In this case, the only horizontal force acting on the box is the force applied parallel to the ground, which has a magnitude of F = 100.0 N. The mass of the box is m = 1.00 kg.
Using Newton's second law, we can rearrange the equation to solve for acceleration:
a = F / m
Plugging in the values, we get:
a = 100.0 N / 1.00 kg
a = 100.0 m/s^2
Therefore, the linear acceleration magnitude of the box is 100.0 m/s^2.
(b) To find the tension T in the string, we can consider the torque acting on the disk.
The torque produced by the force F applied to the box is equal to the product of the force and the radius of the disk (τ = F * r). The torque causes the disk to rotate.
The torque produced by the tension T in the string is equal to the product of the tension and the radius of the disk (τ = T * r). The torque by the tension acts in the opposite direction to the torque by the force applied to the box.
Since the disk is rotating, the net torque acting on the disk is zero:
0 = τ(force) - τ(tension)
0 = F * r - T * r
T * r = F * r
So, the tension T in the string is equal to the force F applied to the box.
Therefore, the tension T in the string is 100.0 N.
(c) If the disk is replaced with a hollow thin-walled cylinder of the same mass and diameter, the acceleration in part (a) will be the same.
This is because the hollow thin-walled cylinder has the same mass and diameter as the original disk. Therefore, the moment of inertia (a property related to rotational motion) of the cylinder will also be the same. As a result, the torque produced by the applied force and the tension in the string will be the same, leading to the same linear acceleration magnitude for the box.