The following are the answers to the question presented:
a. magnitude of the radial acceleration = 1.25m/s² inwardly directed
b. tangential acceleration = 0.400m/s²
c. total acceleration = 72.25 degrees
I am hoping that these answers have satisfied your queries and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.
Answer:
Part a)
[tex]a_r = 1.25 m/s^2[/tex]
Part b)
[tex]a_t = 0.4 m/s^2[/tex]
Part c)
[tex]a_{net} = 1.31 m/s^2[/tex]
Explanation:
initial speed of the turn table at t = 0 is given as
[tex]v_i = 0[/tex]
after t = 1.75 s the speed is given as
[tex]v_f = 0.700 m/s[/tex]
now the tangential acceleration is given as
[tex]a_t = \frac{v_f - v_i}{t}[/tex]
[tex]a_t = \frac{0.700 - 0}{1.75}[/tex]
[tex]a_t = 0.4 m/s^2[/tex]
Part a)
Now the speed of the disc after t = 1.25 s is given as
[tex]v_f = v_i + at[/tex]
here we will have
[tex]v_f = 0 + (0.4)(1.25)[/tex]
[tex]v_f = 0.5 m/s[/tex]
now radial acceleration is given as
[tex]a_r = \frac{v^2}{R}[/tex]
[tex]a_r = \frac{0.5^2}{0.20}[/tex]
[tex]a_r = 1.25 m/s^2[/tex]
Part b)
tangential acceleration is given as
[tex]a_t = 0.4 m/s^2[/tex]
Part c)
total acceleration is given as
[tex]a_{net} = \sqrt{a_r^2 + a_t^2}[/tex]
[tex]a_{net} = \sqrt{1.25^2 + 0.4^2}[/tex]
[tex]a_{net} = 1.31 m/s^2[/tex]
