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Newton's Laws with Calculus | Study Mondo
Topics / Dynamics / Newton's Laws with Calculus Newton's Laws with Calculus Force, mass, and acceleration relationships using derivatives and integrals
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Newton's Second Law
Vector form:
F ⃗ n e t = m a ⃗ = m d v ⃗ d t = m d 2 r ⃗ d t 2 \vec{F}_{net} = m\vec{a} = m\frac{d\vec{v}}{dt} = m\frac{d^2\vec{r}}{dt^2} F
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▾ 📌 Related Topics in Dynamics❓ Frequently Asked QuestionsWhat is Newton's Laws with Calculus?▾ Force, mass, and acceleration relationships using derivatives and integrals
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Component form:
F x = m d v x d t , F y = m d v y d t , F z = m d v z d t F_x = m\frac{dv_x}{dt}, \quad F_y = m\frac{dv_y}{dt}, \quad F_z = m\frac{dv_z}{dt} F x = m d t d v x , F y = m d t d v y , F z = m d t d v z
Impulse-Momentum Theorem From Newton's second law:
F ⃗ = d p ⃗ d t \vec{F} = \frac{d\vec{p}}{dt} F = d t d p
where momentum p ⃗ = m v ⃗ \vec{p} = m\vec{v} p = m v .
Integrating both sides:
∫ t 1 t 2 F ⃗ d t = ∫ t 1 t 2 d p ⃗ d t d t = p ⃗ 2 − p ⃗ 1 = Δ p ⃗ \int_{t_1}^{t_2} \vec{F} \, dt = \int_{t_1}^{t_2} \frac{d\vec{p}}{dt} \, dt = \vec{p}_2 - \vec{p}_1 = \Delta \vec{p} ∫ t 1 t 2 F d t = ∫ t 1 t 2 d t p 2 − p 1 = Δ p
Impulse:
J ⃗ = ∫ t 1 t 2 F ⃗ d t = Δ p ⃗ \vec{J} = \int_{t_1}^{t_2} \vec{F} \, dt = \Delta \vec{p} J = ∫ t 1 t 2 F d t = Δ p
Variable Force Example If F ( t ) = F 0 cos ( ω t ) F(t) = F_0\cos(\omega t) F ( t ) = F 0 cos ( ω t ) for time 0 0 0 to T = π / ( 2 ω ) T = \pi/(2\omega) T = π / ( 2 ω ) :
J = ∫ 0 T F 0 cos ( ω t ) d t = F 0 ω sin ( ω t ) ∣ 0 T = F 0 ω J = \int_0^T F_0\cos(\omega t) \, dt = \frac{F_0}{\omega}\sin(\omega t)\Big|_0^T = \frac{F_0}{\omega} J = ∫ 0 T F 0 cos ( ω t ) d t = ω F 0 sin ( ω t ) ω F 0
Variable Mass Systems For systems where mass changes with time (rockets):
F ⃗ e x t = d p ⃗ d t = d ( m v ⃗ ) d t = m d v ⃗ d t + v ⃗ d m d t \vec{F}_{ext} = \frac{d\vec{p}}{dt} = \frac{d(m\vec{v})}{dt} = m\frac{d\vec{v}}{dt} + \vec{v}\frac{dm}{dt} F e x t = d t d p = d t d ( m v ) = m d t d v + v d t d m
Rocket equation:
m d v d t = − v r e l d m d t + F e x t m\frac{dv}{dt} = -v_{rel}\frac{dm}{dt} + F_{ext} m d t d v = − v re l d t d m + F e x t
where v r e l v_{rel} v re l is exhaust velocity relative to rocket.
For no external forces and integrating:
Δ v = v r e l ln m 0 m f \Delta v = v_{rel}\ln\frac{m_0}{m_f} Δ v = v re l ln m f m 0
(Tsiolkovsky rocket equation)
Drag Forces Linear drag: F d = − b v F_d = -bv F d = − b v
m d v d t = − b v m\frac{dv}{dt} = -bv m d t d v = − b v
Solution: v ( t ) = v 0 e − b t / m v(t) = v_0e^{-bt/m} v ( t ) = v 0 e − b t / m
Quadratic drag: F d = − c v 2 F_d = -cv^2 F d = − c v 2
m d v d t = − c v 2 m\frac{dv}{dt} = -cv^2 m d t d v = − c v 2
Separating variables:
∫ v 0 v d v ′ v ′ 2 = − c m ∫ 0 t d t ′ \int_{v_0}^v \frac{dv'}{v'^2} = -\frac{c}{m}\int_0^t dt' ∫ v 0 v v ′2 d v ′ = − m c ∫ 0 t d t ′
v ( t ) = v 0 1 + c v 0 t m v(t) = \frac{v_0}{1 + \frac{cv_0t}{m}} v ( t ) = 1 + m c v 0 t v 0
Falling with Air Resistance Equation of motion:
m d v d t = m g − b v m\frac{dv}{dt} = mg - bv m d t d v = m g − b v
Terminal velocity: v t = m g b v_t = \frac{mg}{b} v t = b m g
Rewrite as:
d v d t = g − b m v = b m ( v t − v ) \frac{dv}{dt} = g - \frac{b}{m}v = \frac{b}{m}(v_t - v) d t d v = g − m b v = m b ( v t − v )
Solution:
v ( t ) = v t ( 1 − e − b t / m ) v(t) = v_t(1 - e^{-bt/m}) v ( t ) = v t ( 1 − e − b t / m )
Forces in Polar Coordinates Newton's second law in polar coordinates:
Radial direction:
F r = m ( r ¨ − r θ ˙ 2 ) F_r = m(\ddot{r} - r\dot{\theta}^2) F r = m ( r ¨ − r θ ˙ 2 )
Tangential direction:
F θ = m ( r θ ¨ + 2 r ˙ θ ˙ ) F_{\theta} = m(r\ddot{\theta} + 2\dot{r}\dot{\theta}) F θ = m ( r θ ¨ + 2 r ˙ θ ˙ )
Example: Central Force For motion under central force (depends only on r r r ):
F θ = 0 F_{\theta} = 0 F θ = 0 , so:
r θ ¨ + 2 r ˙ θ ˙ = 0 r\ddot{\theta} + 2\dot{r}\dot{\theta} = 0 r θ ¨ + 2 r ˙ θ ˙ = 0
d d t ( r 2 θ ˙ ) = 0 \frac{d}{dt}(r^2\dot{\theta}) = 0 d t d ( r 2 θ ˙ ) = 0
This means r 2 θ ˙ = L / m r^2\dot{\theta} = L/m r 2 θ ˙ = L / m is constant (angular momentum conservation).
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