Question

A thin, light string is wrapped around the rim of a 3.75-kg , solid, uniform disk (similar to a puck on ice without friction) that is 35.0 cm in diameter. A person pulls on the string with a constant force of 101.5 N tangent to the disk, as shown in the figure below (Figure 1). The disk is not attached to anything and is free to move and turn. Find the linear acceleration of its center of mass.

Answer 1

The linear acceleration of the solid uniform disk's center of mass, when pulled by a string with a force of 101.5 N, is 27.0667 m/s².

Step-by-step explanation:

The student asked to find the linear acceleration of the center of mass for a solid uniform disk being pulled by a string. Since there is no friction, only the force from the string and the inertia of the disk are involved. The linear acceleration (a) can be found using Newton's second law of motion, F = ma, where F is the force applied, m is the mass of the disk, and a is the acceleration of the disk.

The mass of the disk is given as 3.75 kg and the force applied is 101.5 N. Thus, the linear acceleration a can be calculated as follows:

a = F / m

a = 101.5 N / 3.75 kg

a = 27.0667 m/s²

Answer 2

Γ = I α this is the rotational equivalent of F = M a for linear motion

The puck has a radius of .35 m / 2 = .175 m

The mass of the puck is 3.75 kg

I = 1/2 M R^2 for a uniform disc

We can use F = M a here to describe the center of mass motion

a = F / m = 101.5 N / 3.75 kg = 27.1 m/s^2 because there are no other external forces acting on the mass