A uniform drawbridge 8.50 m long is attached to the roadway by a frictionless hi

Question

A uniform drawbridge 8.50 m long is attached to the roadway by a frictionless hinge at one end, and it can be raised by a cable attached to the other end. The bridge is at rest, suspended at 70.0 above the horizontal, when the cable suddenly breaks.

Find the angular acceleration of the drawbridge just after the cable breaks. (Gravity behaves as though it all acts at the center of mass.)

Could you use the equation =0+t to calculate the angular speed of the drawbridge at a later time? Explain why.

What is the angular speed of the drawbridge as it becomes horizontal?

Explanation / Answer

For part 1, use we have the following identities of torque:

= Fr = I

F = (mg)cos for the component of the force perpendicular to the lever arm
r = d/2
I = (1/3)md^2 for rectangular plate rotating about its side

Given: d = 8.50 and = 70.0

Now place these equations in place for torque.

= mg(d/2)cos = (1/3)(md^2)

Notice that masses cancel out as well as one d, leaving:

(1/2)gcos = (1/3)d

Solve for

= ((3/2) g cos) / d = 1.04 rad/s^2

For part 2, we cannot use this angular acceleration as it does not remain constant. Therefore, use energy as it remains constant.

PE(initial) = K(final)

K(final) = (1/2)I^2 for rotational energy
I = (1/3)md^2
h(initial) = 4.5*sin(70) measured from the center of mass.

mgh = (1/2)I^2

mgh = (1/6)(md^2)(^2) *Note how again masses cancel out.

Solve for :

=[ (6(4.5sin(70))/ d^2]^(1/2) = 0.513 rad/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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