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

We can start by finding the acceleration of the system. Taking the direction of motion of the heavier mass as positive, we have:
$$\sum F = m_{total}a$$
$$T - m_1g + m_2g -\frac{I}{r^2}a = m_{total}a$$
where $m_{total} = m_1 + m_2$
Thus,
$$a = \frac{m_2 - m_1}{m_{total}+\frac{I}{r^2}}g$$
Substituting the given values, we get:
$$a = \frac{710	ext{ g}-385	ext{ g}}{(710	ext{ g}+385	ext{ g})+\frac{0.0125	ext{ kg m}^2}{(0.15	ext{ m})^2}} 	imes 9.8 	ext{ m/s}^2 \approx 2.56 	ext{ m/s}^2$$
Now, we can use this acceleration to find the tension on the side of the 710-g tile:
$$T - m_1g = m_1a$$
$$T = m_1g + m_1a = m_1(g+a)$$
Substituting the given values, we get:
$$T = 385	ext{ g}(9.8	ext{ m/s}^2 + 2.56	ext{ m/s}^2) \approx 5.59 	ext{ N}$$
Therefore, the answer is C.

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 $0.0125 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$