A uniform disk with radius 0.20 m is pivoted at its center about a horizontal fr
ID: 1272890 • Letter: A
Question
A uniform disk with radius 0.20 m is pivoted at its center about a horizontal frictionless axle. The moment of inertia of the disk about an axis at the axle is 40 kg m2. The disk is initially at rest and then a constant force of F = 80 N is applied tangent to the rim of the disk. What is the magnitude of the angular acceleration of the disk? What is the magnitude v of the tangential velocity of a point on the rim of the disk 5.0 s after the force starts to be applied? A uniform beam is 12.0 m long and weighs 700 N. The lower end of the beam is attached to a vertical wall by a frictionless hinge. A light horizontal rope is attached to the upper end of the beam and holds the beam at an angle of 53degree above the horizontal. For an axis at the hinge, what is the magnitude of the torque produced by the weight of the beam? What is the tension in the rope?Explanation / Answer
6.
Radius = 0.2 m
Inertia = 40
angular acceleration = Torque / inertia
= force * R(perpendicular) / 40
= 80 *0.2 /40
= 0.4
b) W(angular vel) = w(initial ang. velocity ) + angular acceleration * time
W = 0 + 0.4*5 = 2 m/s
7)
Torque = Force * radius (perpendicular)
= 700 * 6* sin37
= 700 *6 * 3/5
= 2520 N m
b)
balancing forces perpndicular to the hinge force in the rod
T cos37 = 700 sin37
(T = tension in rope)
T*4/5 = 700 *3/5
T = 525 N
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