F.a

Classical Mechanics Level 3

A force of constant magnitude F F starts acting on the end A A of a uniform rod A B AB of mass M M and length L L in gravity-free space. The force always remains perpendicular to the rod, even as it moves.

If a A → \overrightarrow{a_{A}} is the acceleration of point A , A, then what is the value of the dot product F → ⋅ a A → \overrightarrow{F} \cdot \overrightarrow{a_{A}} at any later time?

0 0 F 2 M \frac{F^2}{M} 2 F 2 M \frac{2F^2}{M} 3 F 2 M \frac{3F^2}{M} 4 F 2 M \frac{4F^2}{M}

Vilakshan Gupta by Vilakshan Gupta , 19, India

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1 solution

Steven Chase
Nov 9, 2018

Position at point of force application - relative to center of mass:

x = x c + L 2 c o s θ y = y c + L 2 s i n θ \large{x = x_c + \frac{L}{2} cos \theta \\ y = y_c + \frac{L}{2} sin \theta}

Velocity:

x ˙ = x c ˙ − L 2 s i n θ θ ˙ y ˙ = y c ˙ + L 2 c o s θ θ ˙ \large{\dot{x} = \dot{x_c} - \frac{L}{2} sin \theta \, \dot{\theta} \\ \dot{y} = \dot{y_c} + \frac{L}{2} cos \theta \, \dot{\theta}}

Acceleration:

x ¨ = x c ¨ − L 2 s i n θ θ ¨ − L 2 c o s θ θ ˙ 2 y ¨ = y c ¨ + L 2 c o s θ θ ¨ − L 2 s i n θ θ ˙ 2 \large{\ddot{x} = \ddot{x_c} - \frac{L}{2} sin \theta \, \ddot{\theta} - \frac{L}{2} cos \theta \, \dot{\theta}^2 \\ \ddot{y} = \ddot{y_c} + \frac{L}{2} cos \theta \, \ddot{\theta} - \frac{L}{2} sin \theta \, \dot{\theta}^2}

Angular Acceleration :

τ = I θ ¨ F L 2 = 1 12 M L 2 θ ¨ θ ¨ = 6 F M L \large{\tau = I \ddot{\theta} \\ F \frac{L}{2} = \frac{1}{12} M L^2 \, \ddot{\theta} \\ \ddot{\theta} = \frac{6 F}{M L}}

Vector Force :

F x = − F s i n θ F y = F c o s θ \large{F_x = -F \, sin\theta \\ F_y = F \, cos\theta}

Acceleration of Center of Mass :

x c ¨ = − F s i n θ M y c ¨ = F c o s θ M \large{\ddot{x_c} = \frac{-F \, sin\theta}{M} \\ \ddot{y_c} = \frac{F \, cos\theta}{M}}

Substituting angular acceleration and center of mass acceleration into x and y acceleration equations:

x ¨ = − F s i n θ M − 3 F M s i n θ − L 2 c o s θ θ ˙ 2 = − 4 F M s i n θ − L 2 c o s θ θ ˙ 2 y ¨ = F c o s θ M + 3 F M c o s θ − L 2 s i n θ θ ˙ 2 = 4 F M c o s θ − L 2 s i n θ θ ˙ 2 \large{\ddot{x} = \frac{-F \, sin\theta}{M} - \frac{3 F}{M} sin \theta - \frac{L}{2} cos \theta \, \dot{\theta}^2 \\ = -\frac{4 F}{M} sin \theta - \frac{L}{2} cos \theta \, \dot{\theta}^2 \\ \ddot{y} = \frac{F \, cos\theta}{M} + \frac{3 F}{M} cos \theta - \frac{L}{2} sin \theta \, \dot{\theta}^2 \\ = \frac{4 F}{M} cos \theta - \frac{L}{2} sin \theta \, \dot{\theta}^2 }

Dot product between vector force and acceleration at application point:

F x x ¨ + F y y ¨ = 4 F 2 M s i n 2 θ + F L 2 s i n θ c o s θ θ ˙ 2 + 4 F 2 M c o s 2 θ − F L 2 s i n θ c o s θ θ ˙ 2 = 4 F 2 M s i n 2 θ + 4 F 2 M c o s 2 θ = 4 F 2 M \large{F_x \, \ddot{x} + F_y \, \ddot{y } \\ = \frac{4 F^2}{M} sin^2 \theta + \frac{F L}{2} sin \theta \, cos \theta \, \dot{\theta}^2 + \frac{4 F^2}{M} cos^2 \theta - \frac{F L}{2} sin \theta \, cos \theta \, \dot{\theta}^2 \\ = \frac{4 F^2}{M} sin^2 \theta + \frac{4 F^2}{M} cos^2 \theta \\ = \frac{4 F^2}{M}}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Sir, can we solve this problem by IAR method ?

Vilakshan Gupta - 2 years, 6 months ago

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