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
Answers
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:
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.
(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;
(we have taken negative sign because it is slowing down)
Now radial acceleration(ar),
(2)
By substituting the value of r and ω in equation (2) we get;
Now total acceleration of point A will be given as,
(3)
by substituting the value of at and ar in equation (3) we get;
(b)
Just like point A,the acceleration of point B will also have two parts.
By using equations (1) and (2) we have;
Total acceleration of point B,
The relative acceleration,
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.