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umka21 [38]
3 years ago
9

To understand and apply the formula τ=Iα to rigid objects rotating about a fixed axis. To find the acceleration a of a particle

of mass m, we use Newton's second law: F net=ma , where F net is the net force acting on the particle.To find the angular acceleration α of a rigid object rotating about a fixed axis, we can use a similar formula: τnet=Iα, where τnet=∑τ is the net torque acting on the object and I is its moment of inertia.
Part A:
Assume that the mass of the swing bar, is negligible. Find the magnitude of the angular acceleration α of the seesaw.
Express your answer in terms of some or all of the quantities m1, m2, l, as well as the acceleration due to gravity g.
Part B:
Now consider a similar situation, except that now the swing bar itself has mass mbar.Find the magnitude of the angular acceleration α of the seesaw.
Express your answer in terms of some or all of the quantities m1, m2, mbar, l, as well as the acceleration due to gravity g.
Physics
1 answer:
Alenkinab [10]3 years ago
7 0

Answer:

Hello your question is incomplete attached below is the missing part of the question

In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these situations, two objects of masses m1 and m2 are attached to a seesaw. The seesaw is made of a bar that has length l and is pivoted so that it is free to rotate in the vertical plane without friction. Assume that the pivot is attached tot he center of the bar.

You are to find the angular acceleration of the seesaw when it is set in motion from the horizontal position. In all cases, assume that m1>m2.I

answer : part A = 2*[(M1 - M2)/(M1 + M2)]*g/L

              part A = attached below

Explanation:

Part A :

Assuming that mass of swing is negligible

α = T/I

where ; T = torque, I = inertia,

hence T =  L/2*9*(M1 - M2)

also;  I = M1*(L/2)^2 + M2*(L/2)^2=  ( M1 + M2) * (L/2)^2

Finally the magnitude of the angular acceleration α

α = 2*[(M1 - M2)/(M1 + M2)]*g/L

Part B attached below

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Answer :

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(b) The acceleration is, -2.33m/s^2

Explanation :

By the 1st equation of motion,

v=u+at     ...........(1)

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By the 2nd equation of motion,

s=ut+\frac{1}{2}at^2     ...........(3)

where,

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Now substitute equation 2 in 3, we get:

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By solving the term, we get:

a=-2.33m/s^2

The acceleration is, -2.33m/s^2

Now we have to calculate the initial velocity.

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u=-6.35a

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