A ball of mass m = 0.1kg falls from rest under the influence of gravity (on earth) in a medium that provides resistance that is proportional to its velocity. For a velocity of 0.2 m/s, the resistance force on the object is measured to be 1N.
How long does the ball take to reach half of its terminal velocity?
I found the terminal velocity to be -.0196 m/s but I am not sure how to model the velocity equation to solve for time t when the velocity is half of my terminal velocity. Thank you!
On
The net force acting on the ball is:$$F=(\text{Force due to Gravity/ Weight)}-(\text{Resistive Force})$$ Now weight is $ mg=0.1\times 9.8=0.98 N$ and resistive force is $F_r\propto v$ or $F_r=kv=5v$ (by given condition). So, the net force becomes: $$F = 0.98-5v$$ By Newton's second law of motion, $F=ma=0.1\displaystyle\frac{dv}{dt}$. Hence $$0.1\frac{dv}{dt}= 0.98-5v$$ $$\frac{dv}{dt}=9.8-50v$$ $$\frac{1}{9.8-50v}dv=dt$$ Integrating both sides, $$-\frac1{50}\ln(9.8-50v)=t+c^*$$ $$9.8-50v = ce^{-50t}$$ $$v=\frac{9.8-ce^{-50t}}{50}$$ Since the ball starts at rest so $v(0)=0.$ Hence $$\frac{9.8-c}{50}=0\ \ \ \ \ \Rightarrow \ \ \ c=9.8$$ So $$v= \frac{9.8-9.8e^{-50t}}{50}$$ The terminal velocity is found by putting $t\rightarrow \infty$ i.e. $v_t=\displaystyle\frac{9.8}{50}=0.196$.
Now, we have to find the time $t$ for which $v(t)=\displaystyle\frac{v_t}2=0.098$. So $$\frac{9.8-9.8e^{-50t}}{50} = 0.098$$ $$9.8-9.8e^{-50t} = 4.9$$ $$9.8e^{-50t} = 4.9$$ $$e^{-50t}=0.5$$ $$t = \frac{\ln0.5}{-50}=0.01386 \text{ seconds}$$
By Newton's second law (rate of change of momentum is equal to net force applied),
$$m \frac{dv}{dt} = mg - \alpha v$$; $\alpha $ is a constant, $\alpha v$ is the resisitive force.
Integrate w.r.t. time to get $v$ in terms of $t$.
$v|_{t = 0} = 0$ and $v|_{t \to \infty} = v_t$, terminal velocity is attained when $\frac{dv}{dt} = 0$, i.e. $v_t = \frac{mg}{\alpha}$
$m \int_{0}^{\frac{v_t}{2}} \frac{dv}{mg - \alpha v} = \int_{0}^{t_0}dt\quad$; $t_0$ is the required time.