Uniformly accelerated circular motion • pickedshares

Task

A mass point moves with a constant angular acceleration c₁ on a circular path of radius R. How big must the angular acceleration be so that the mass point comes to a standstill after one revolution?

Given:

$\ddot \phi = c_1$

$\tag{IC 1} \dot \phi (t=0) = \omega_0$

$\tag{IC 2} \phi (t=0) = \phi_0$

Solution

A solution video will be published soon on this channel.

Since it is a circular movement with a constant radius, the reduced consideration of the angle of rotation, the angular velocity and the angular acceleration is sufficient. The function is obtained by integrating the angular acceleration

$\tag{1} \int \ddot \phi dt \rightarrow \dot \phi (t) = c_1 \cdot t + c_2$

With the initial condition IC1 c₂ can be solved.

$\tag{2} \dot \phi (t = 0) = \omega_0$

$\tag{3} c_2 = \omega_0$

$\tag{4} \dot \phi (t) = c_1 \cdot t + \omega_0$

By integrating the angular velocity it follows

$\tag{5} \int \dot \phi dt \rightarrow \phi (t) = \frac{1}{2} c_1 \cdot t^2 + \omega_0 t + c_3$

Using the initial condition IC2 we can find c₃:

$\tag{6} \phi (t = 0) = \phi_0$

$\tag{7} c_3 = \phi_0$

$\tag{8} \phi (t) = \frac{1}{2} c_1 \cdot t^2 + \omega_0 t + \phi_0$

The point in time at which the movement comes to a standstill is currently unknown and is referred to below as T. At time T, a complete revolution has taken place

$\tag{9} \phi (t = T) = 2 \pi$

$\tag{10} 2 \pi = \frac{1}{2} c_1 T^2 + \omega_0 T + \phi_0$

and the angular velocity is zero.

$\tag{11} \dot \phi (t = T) = 0$

$\tag{12} 0 = c_1 T + \omega_0$

$\tag{13} c_1 = - \frac{\omega_0}{T}$

The determined c₁ will be used in equation 10 and solved for T.

$\tag{14} T = \frac{2 \cdot (2 \pi - \phi_0)}{\omega_0}$

T used in equation 13 yields the angular acceleration we are looking for.

$\tag{15} c_1 = - \frac{\omega_0^2}{2 \cdot (2 \pi - \phi_0)}$