A uniform disk with mass 40.1 and radius 0.280 is pivoted at its center about a

Question

A uniform disk with mass 40.1 and radius 0.280 is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 28.5 is applied tangent to the rim of the disk.

a) What is the magnitude of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.320 revolution?

b) What is the magnitude of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.320 revolution?

Explanation / Answer

use torque = I alpha where I is the moment of inertia and alpha is the angular acceleration the moment of inertia of a soid disk is 1/2 MR^2 the torque is F x R since the force is applied tangentially to the rim of the disk, so we have F R = 1/2 MR^2 alpha solve for alpha: alpha = 2 F/M R = 2*28.5N/(35.2 kg * 0.2m) = 8.10 rad/s/s 0.34 rev = 2.14 rad find angular velocity from wf^2 = w0^2 + 2 alpha theta wf, w0 = final, initial ang vel alpha = ang accel = 8.10rad/s/s theta = angular displacement = 2.14 rad wf^2=0+2*8.10*2.14 => wf = 5.8 rad/s this is the angular velocity, the linear velocity = w r = 1.18m/s resultant accel = sqrt[a centip^2 + a tangential^2] the linear acceleration = alpha r = 8.1rad/s/s x 0.2m = 1.6 m/s/s a centrip = v^2/r = 1.18^2/0.2 = 6.92m/s/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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