An object with a mass of m = 5.5 kg is attached to the free end of a light strin

Question

An object with a mass of  m = 5.5 kg is attached to the free end of a light string wrapped around a reel of radius R = 0.240 m and mass of M = 3.00 kg. The reel is a solid disk, free to rotate in a vertical plane about the horizontal axis passing through its center as shown in the figure below. The suspended object is released from rest 5.90 m above the floor.

(a) Determine the tension in the string.
N

(b) Determine the magnitude of the acceleration of the object.
m/s2

(c) Determine the speed with which the object hits the floor.
m/s

An object with a mass of m = 5.5 kg is attached to the free end of a light string wrapped around a reel of radius R = 0.240 m and mass of M = 3.00 kg. The reel is a solid disk, free to rotate in a vertical plane about the horizontal axis passing through its center as shown in the figure below. The suspended object is released from rest 5.90 m above the floor. (a) Determine the tension in the string. N (b) Determine the magnitude of the acceleration of the object. m/s2 (c) Determine the speed with which the object hits the floor. m/s

Explanation / Answer

I = 1/2 mR^2

I = 1/ 2 * 3 * 0.24^2

I = 0.0864 Nm^2

Tension T = mg-ma

Torque ? = T*R

? = (mg - ma)*R         (1)

? = I*?

? = I*(a/R)                (2)

equating 1 and 2

a = mg/(I/R^2 + m)

a = 5.5 * 9.81 / (0.0864 / 0.24^2 + 5.5)

a = 7.70 m/s^2

acceleration of the object = 7.70 m/s^2

Tension T = mg-ma

Tension = 5.5 * (9.81 - 7.7)

tension in the string = 11.605 N

v = sqrt(2sa)

v = sqrt(2 * 5.9 * 7.7)

v = 9.532 m/s

speed with which the object hits the floor = 9.532 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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