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-PART 1 - (a) If m1 = m2, will the masses have an angular acceleration? (b) Give

ID: 1978646 • Letter: #

Question

-PART 1 -

(a) If m1 = m2, will the masses have an angular acceleration?

(b) Given particular values of L1 and L2, does the angular acceleration depend only on m1 / m2 ? (If it depends on the actual values of m1 and m2, put 'no'.)

(c) Is there a net (non-zero) torque?

(d) Given particular values of m1, m2, and L1, is it always possible to choose L2 (with L2 > L1) such that the masses have no angular acceleration?

-PART2-

If L1 = 0.670 m, L2 = 1.70m, m1 = 5.10 kg, and m2 = 2.15 kg, what is the angular acceleration of the beam?

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Explanation / Answer

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A beam is supported only at one point, called the pivot point, as shown in the diagram. A block with mass m1 sits at the left end of the beam, a distance L1 from the pivot point. A block with mass m2 sits at the right end of the beam, a distance L2 from the pivot point. L2 > L1. Calculate all torques about the pivot point, remembering that positive is anti-clockwise.
If L1 = 0.930 m, L2 = 1.00m, m1 = 4.25 kg, and m2 = 2.30 kg, what is the angular acceleration of the beam?

ANSWER:

Torque = force * perpendicular distance
Force = weight = m* 9.8

Torque left side = m1 * 9.8 * L1, anti clockwise
Torque right side = -(m2 * 9.8 * L2), clockwise

Torque = (m1 * 9.8 * L1) – -(m2 * 9.8 * L2)

If L1 = 0.930 m, L2 = 1.00m, m1 = 4.25 kg, and m2 = 2.30 kg, what is the angular acceleration of the beam?


Torque = (4.25 * 9.8 * 0.930) – (2.30 * 9.8 * 1.00) = 16.1945 (anti-clockwise)

Torque = moment of Inertia * angular acceleration

Moment of Inertia = mass * distance from pivot point
Moment of Inertia = (4.25 * 0.930) + (2.30 * 1.00) = 6.2525

16.1945 = 6.2525 *
= 16.1945 ÷ 6.2525 = 2.59 radians per second