A uniform disk with mass 39.5kg and radius 0.250 m is pivoted at its center abou

Question

A uniform disk with mass 39.5kg and radius 0.250 m is pivoted at its center about a horizontal, frictionless axle that is stationary. the disk is initially at rest, and then a constant force of 34.5N is applied tangent to the rim of the disk.

Part A

What is the magnitude v of the tangential velocity of a point on the rim of the disk after the disk had turned through 0.150 revolution?
-Express your answer with appropriate units


Part B

What is the magnitude a of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.150 revolution
-Express your answer with the appropriate units.

Explanation / Answer

a)

I = 0.5 m R2 = 0.5 * 39.5 *0.25*0.25 = 1.234375 Kg.m2

= I = F R

= F R/I = 34.5 * 0.25/1.234375 = 6.9873 rad/s2

2 = 2

= 0.150 * 2 * = 0.94248 rad

2 = 2 = 2 * 6.9873 * 0.94248 = 13.171

= 3.62916 rad/s

v = R = 0.25 * 3.62916 = 0.907 m/s

b)

a = R 2 = 0.25 * 3.62916 * 3.62916 = 3.29 m/s2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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