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vovangra [49]
3 years ago
15

A child is pushing a merry-go-round. the angle through which the merry-go-round has turned varies with time according to θ(t)=γt

+βt3, where γ= 0.422 rad/s and β= 1.35×10−2 rad/s3.

Physics
1 answer:
Andreyy893 years ago
3 0
I attached the missing part of the question.
Part A
Angular velocity is simply the rate at which angle changes, in other words, it is the first derivative of the angle function with respect to time:
\theta (t)=\gamma t+\beta t^3\\ w(t)=\frac{d\theta(t)}{dt}\\ w(t)=\gamma +3\beta t^2
Part B
Initial value is the value at t=0. To find initial value we simply plug t=0 into the equation.
w(0)=\gamma +3\beta\cdot 0^2\\
w(0)=\gamma\\
w_0=\gamma= 0.422\frac{rad}{s}
Part C
To find these values we simply need to plug in t=5 in the equation.
w(5)=\gamma +3\beta\cdot 5^2\\
w(5)=\gamma +75\cdot \beta=0.422+75 \cdot 1.35\cdot10^{-2}=1.43\frac{rad}{s}
Part D
Average angular velocity is total angular displacement divided by time:
w_{av}=\frac{\Delta \theta}{\Delta t}
\Delta \theta=\theta(5)-\theta(0)=\gamma\cdot5+\beta 5^3\\
\Delta \theta=0.422\cdot5+125\cdot1.35\cdot10^{-2}=3.7975 $rad
The average angular velocity is:
w_{av}=\frac{\Delta \theta}{\Delta t}=\frac{3.7975 }{5}=0.7595\frac{rad}{s}

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Q_{enc} = \int\limits^R_0 {\rho(r)h} \, dr = \alpha \int\limits^R_0 {(1-r/R)h} \, dr = \alpha h(r - \frac{r^2}{2R})\left \{ {{r=R} \atop {r=0}} \right. = \alpha h(R - \frac{R^2}{2R}) = \alpha h\frac{R}{2} \\E2\pi rh = \frac{\alpha Rh}{2\epsilon_0}\\E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}

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