Answer:
So the acceleration of the child will be [tex]8.05m/sec^2[/tex]
Explanation:
We have given angular speed of the child [tex]\omega =1.25rad/sec[/tex]
Radius r = 4.65 m
Angular acceleration [tex]\alpha =0.745rad/sec^2[/tex]
We know that linear velocity is given by [tex]v=\omega r=1.25\times 4.65=5.815m/sec[/tex]
We know that radial acceleration is given by [tex]a=\frac{v^2}{r}=\frac{5.815^2}{4.65}=7.2718m/sec^2[/tex]
Tangential acceleration is given by
[tex]a_t=\alpha r=0.745\times 4.65=3.464m/sec^[/tex]
So total acceleration will be [tex]a=\sqrt{7.2718^2+3.464^2}=8.05m/sec^2[/tex]
The magnitude of the linear acceleration of the child is mathematically given as
a=8.05m/sec^2
Question Parameters:
A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 rad/s2.
Generally the equation for the linear velocity is mathematically given as
v=wr
Therefore
v=1.25*4.65
v=5.815
radial acceleration is given by
a=v^2/r
Hence
a=5.815/4.65
a=7.2718
Tangential acceleration is
a_t=\alpha r
a_t=0.745*4.65
a_t=3.464m/sec
Hence, total acceleration will be
[tex]a=\sqrt{7.2718^2+3.464^2}[/tex]
a=8.05m/sec^2
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