What will be the effect on a rigid body if two unlike unequal parallel forces act on the body? Will the body translate or rotate or both? If they rotate, with respect too which point will they rotate?
[Physics] What will be the effect on a rigid body if two unlike unequal parallel forces act on the body
forcesnewtonian-mechanicsreference framesrotational-dynamicstorque
Related Solutions
In the diagram shown there are external forces. The reaction from the ground is such that the velocity on the contact point is horizontal only. This means the center of rotation is somewhere along the vertical line passing through the contact point.
But where? This depends on the friction condition at the contact point. Consider the general case below:
In order to find the equations of motion we need to establish how this thing moves. We describe this with the variable $x$ for horizontal position of the center A and the angle $\theta$ the part makes with the vertical direction.
Consider sequentially the velocities of points A, B and C when $x$ and $\theta$ vary only.
$$\begin{aligned} \mathbf{v}_A & = \pmatrix{\dot{x} \\ 0} \\ \mathbf{v}_B & = \pmatrix{\dot{x} + r \dot{\theta} \\ 0} \\ \mathbf{v}_C & = \pmatrix{\dot{x} + h \,\dot{\theta} \cos \theta \\ h \, \dot{\theta} \sin \theta} \end{aligned}$$
where $\dot{x}$ and $\dot{\theta}$ are the first time derivatives.
Take the time derivative of the velocity at C to get the acceleration of the center of mass
$$ \mathbf{a}_C = \pmatrix{ \ddot{x} + h\,\ddot{\theta} \cos\theta - h\,\dot{\theta}^2 \sin\theta \\ h\,\ddot{\theta} \sin \theta + h\,\dot{\theta}^2 \cos\theta} $$
Now let's look at the equations of motion by considering all the forces acting on the body.
The equations of motion have to consider the balance of moments about the center of mass (point C) to be valid.
$$ \begin{aligned} \pmatrix{F \\ N - m g } & = m \mathbf{a}_C & & \mbox{sum of forces} \\ -(h\sin\theta) N + (r-h \cos\theta) F & = I_C \ddot{\theta} & & \mbox{sum of moments} \end{aligned} $$
There are three equations and 4 unknowns ($N$, $F$, $\ddot{x}$, $\ddot{\theta}$). To solve them you need an expression descripting the contact condition. Here are the three scenarios
No Slipping - Solve with $\dot{x} + r \dot{\theta}=0$, or $\ddot{x}=-r\,\ddot{\theta}$ for $$\begin{aligned} \ddot{\theta} &= - \frac{h\,m (g+r \dot{\theta}^2) \sin\theta}{I_C+m (r^2+h^2-2 h r \cos\theta)} \\ \ddot{x} & = \frac{r\,h\,m (g+r \dot{\theta}^2) \sin\theta}{I_C+m (r^2+h^2-2 h r \cos\theta)} \end{aligned}$$ The center of rotation height above the ground is $c = r + \frac{\ddot{x}}{\ddot{\theta}} = 0$ so the body is always rotating about point B. This can be confirmed by the fact that the parallel axis theorem in the denominator of $\ddot{\theta}$ contains the distance between B and C.
Zero Friction - Solve with $F=0$ for $$\begin{aligned} \ddot{\theta} & = - \frac{h\, m \sin\theta (g+h \dot{\theta}^2 \cos\theta)}{I_C +m h^2 \sin^2 \theta} \\ \ddot{x} & = \frac{h^2 m \sin\theta\cos\theta (g+h \dot{\theta}^2 \cos\theta)}{I_C + m h^2 \sin^2 \theta} + h \dot{\theta}^2 \sin\theta \end{aligned} $$ The center of rotation height above the ground is $c = r + \frac{\ddot{x}}{\ddot{\theta}}$ which initially (when $\dot{\theta}=0$) is equal to $c = r-h \cos\theta$. So the body rotated about a point along the contact normal with the same height as the center of mass.
- Coulomb Friction - Solve with $F =\pm \mu N$ for $\ddot{x}$, $\ddot{\theta}$ and $N$. I am omitting this solution here for brevity because it is rather complex. The body rotated about a point between the two other solutions depending on the value of $\mu$.
For the particular problem, I assumed no slipping, so set $\theta = \frac{\pi}{2}$, $\dot{\theta}=0$ and $\ddot{x}-r \ddot{\theta} = 0$ to get the center of rotation at the contact point B and the angular acceleration $$\ddot{\theta} =- \frac{h m g}{I_C + m (h^2+r^2)}$$
The key phrase is "with respect to". That means a frame of reference where one of the particles is fixed in space. Since the distance to the other is fixed, any relative motion of the other particle must be circular.
Best Answer
Answer: both.
Now, to determine the point about which they rotate, calculate the linear acceleration (of center of mass) and the angular acceleration (using the torque equation about the center of mass). Then, the acceleration of any point on the body would be the vector sum of linear acceleration and distance times the angular acceleration. The required point is the one for which this vector sum vanishes.