A 385-g tile hangs from one end of a string that goes over a pulley with a moment of inertia of $0.0125 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$ and a radius of 15.0 cm.A mass of 710 g hangs from the other end of the string.When the tiles are released,the larger one accelerates downward while the lighter one accelerates upward.The pulley has no friction in its axle and turns without the string slipping.What is the tension in the string on the side of the 710-g tile?
A)9.77 N
B)6.87 N
C)5.59 N
D)4.41 N
E)3.68 N

Question

Quizplus · Accepted Answer

The tension in the string on the side of the 710-g tile can be found by considering the forces and torques acting on the system. The system includes the pulley and the two masses. The tension difference between the two sides of the string causes the pulley to rotate, and this tension difference must also provide the net force to accelerate both masses. Given the moment of inertia (I) of the pulley, its radius (r), and the masses involved, we can use the rotational form of Newton's second law ($I\alpha = \tau$) and the linear form ($F = ma$) to solve for the tension. The correct tension value accounts for the system's dynamics, including the pulley's inertia and the gravitational forces on the masses.

A 385-G Tile Hangs from One End of a String 0.0125 kg⋅m20.0125 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }0.0125 kg⋅m2

A 385-G Tile Hangs from One End of a String $0.0125 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$