A block of mass m = 2.5 kg is attached to a string that is wrapped around the ci

Question

A block of mass m = 2.5 kg is attached to a string that is wrapped around the circumference of a wheel of radius R = 9.0 cm. The wheel rotates freely about its axis and the string wraps around its circumference without slipping. Initially the wheel rotates with an angular speed omega, causing the block to rise with a linear speed v = 0.30 m/s .(Figure 1) Find the moment of inertia of the wheel if the block rises to a height of h = 6.7 cm before momentarily coming to rest. Express your answer using two significant figures.

Explanation / Answer

Applying

vf^2 - vi^2 = 2 a d

0^2 - 0.30^2 = 2 (a (0.067)

a = - 0.672 m/s^2

acceleration is 0.672 m/s^2 downwards.

for wheel, alpha = a / R = 0.672 / 0.09 = 7.47 rad/s^2

for block:

mg - T = m a

T = m (g - a) = 2.5(9.8 - 0.672) = 26.18 N

for wheel,

torque = I alpha

R T = I alpha

0.09 x 26.18 = I (7.47)

I = 0.315 kg m^2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.