Answer:
[tex]a. \ \gamma +3\beta t^2\\\\b. \ 0.350rad/s\\\\\\c. \omega_i_n_s_t=1.25rad/s\\\\d. w_a_v=0.7rad/s[/tex]
Explanation:
a. Angular velocity,[tex]\omega[/tex] is defined as the rate of change of position of a rotating body with respect to time. so omega,[tex]\omega[/tex]= theta/time and theta =the position angle.
-Angular velocity of the merry-go-round can thus be calculated and expressed as:
[tex]\omega=\frac{d\thata}{dt}=\frac {d(\gamma t+\beta t^3)}{dt}\\=\gamma +3\beta t^3[/tex]
b. From a above, we know that our angular velocity is [tex]\omega=\gamma +3\beta t^3[/tex] ,we can substitute [tex]t=0[/tex] to find our initial velocity.
[tex]\omega_t_=_o=\gamma+3\beta \times (0)^3\\w_t_=_0=0.350rad/s[/tex]
Hence the value of angular velocity at t=0 is 0.350rad/s
c. To calculate the instantaneous angular velocity at t=5.00s, we
[tex]\omega_t_=_5=\gamma +3\beta t^3\\=0.350+ 3\times 0.0120 \times 5^3\\=1.25rad/s[/tex]
Therefore, the instanteneous angular velocity is 1.25rad/s
d. The average angular velocity between t=0 and 2=5s is calculated as
-At t=0,[tex](\theta)=\gamma \times 0+\beta \times 0=0[/tex]
-At time t=5s, [tex]\omega=5\gamma +125\beta\times=3.5rad\\[/tex]
Therefore, the average angular velocity is calculated as
[tex]\omega_a_v=\frac{\bigtriangleup \theta}{\bigtriangleup t}=\frac{\theta_2-\theta_1}{t_2-t_1}\\=\frac{3.5-0}{5-0}\\\\=0.7rad/s[/tex]
Hence the average angular velocity is 0.7rad/s
#instantenous angular velocity does not increase linearly .
