Physics, asked by shashankbrahman, 7 months ago

1. Consider the 'L' shaped bar shown in figure. At the instant shown, the bar is rotating at 4 rad/s and is slowing
down at the rate of 2 rad/s2.
a. Find the acceleration of point A.
b. Find the relative acceleration aB/A of point B with respect to point A.
please give correct answer ​

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Answers

Answered by tlhnizam
0

Answer:

Acceleration of A will have 2 parts, tangentiala  

t

​  

 and radiala  

r

​  

.

a  

t

​  

=α×r=−2×(2)=−4m/s  

2

         -ve sign is because it is slowing down

a  

r

​  

=r×ω  

2

=32m/s  

2

 

total a=  

32  

2

+4  

2

 

​  

=32.25m/s  

2

 

Similarly acceleration of B will have 2 parts.

a  

t

​  

=α×r  

 

=−2×(2  

2

​  

)=−4  

2

​  

m/s  

2

 

a  

r

​  

=r  

 

×ω  

2

=32  

2

​  

m/s  

2

 

total a=  

(32  

2

×2)+(4  

2

×2)

​  

=45.61m/s  

2

 

This is original acceleration of B

But we have to take A as the reference frame which is non inertial.

In this a pseudo acceleration acts in opposite direction.

Hence it is 45.61-35.25= 13.36m/s  

2

Explanation:

Answered by archanajhaa
1

Answer:

The acceleration of point A is 32.25 ms⁻² and relative acceleration aB/A of point B with respect to point A is 13.36 ms⁻².

Explanation:

(a)

Acceleration of A will have two parts i.e. tangential and radial.

a_t=\alpha \times r       (1)

where,

at=tangential acceleration

α=angular acceleration

r=distance from the rotating axis

By substituting the values of α and r in equation (1) we get;

a_t=-2\times 2=-4ms^-^2          (we have taken negative sign because it is slowing down)

Now radial acceleration(ar),

a_r=r\times \omega^2         (2)

By substituting the value of r and ω in equation (2) we get;

a_r=2\times 4^2=32 ms^-^2

Now total acceleration of point A will be given as,

a=\sqrt{a_t^2+a_r^2}            (3)

by substituting the value of at and ar in equation (3) we get;

a_a=\sqrt{4^2+32^2} =32.25ms^-^2

(b)

Just like point A,the acceleration of point B will also have two parts.

By using equations (1) and (2) we have;

a_t=-2\times2\sqrt{2} =-4\sqrt{2} ms^-^2

a_r=2\sqrt{2} \times 4^2=32\sqrt{2} ms^-^2

Total acceleration of point B,

a_b=\sqrt{(4^2\times 2)+(32^2\times 2)} =45.61ms^-^2

The relative acceleration,

a_B_/_A=a_B-a_A=45.61-35.25=13.36ms^-^2

Hence, the acceleration of point A is 32.25 ms⁻² and relative acceleration aB/A of point B with respect to point A is 13.36 ms⁻².

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