Answer:
a=19.8977 m/s²
Explanation:
Given data
ω=3.00 rad/s
r=1.40 m
α=11.0 rad/s²
To find
Acceleration
Solution
As the object moves in a circle so it has tangential acceleration also due to circular motion is has centripetal acceleration
The total acceleration can be found by
[tex]a=\sqrt{(a_{c})^{2}+(a_{T})^{2}}[/tex]
where
at is tangential acceleration
ac is centripetal acceleration
First we need to find centripetal acceleration
so
[tex]a_{c}=rw^{2}[/tex]
put the values or r and ω
[tex]a_{c}=(1.40m)*(3.00rad/s)^{2}\\a_{c}=12.6 m/s^{2}[/tex]
Now for tangential acceleration
[tex]a_{t}=ra\\a_{t}=(1.40m)*(11.0rad/s^{2} )\\a_{t}=15.4 m/s^{2}[/tex]
Put values of ac and at to find total acceleration
So
[tex]a=\sqrt{(a_{t})^{2} +(a_{c})^{2} }\\ a=\sqrt{(15.4m/s^{2} )^{2}+(12.6m/s^{2} )^{2} }\\ a=19.8977m/s^{2}[/tex]
