Question

if you spin a wheel with your hand about an axis located in the middle, how do you calculate the angular acceleration?

Answer 1

Angular acceleration is defined as \[\alpha = {\Delta \omega \over \Delta t}\]If the radius of the wheel is known, the following relations can be used\[\left(\begin{matrix}d_t = \theta r \\ V_t = \omega r \\ a_t = \alpha r\end{matrix}\right)\]where the subscript t indicates the tangential component to the wheel.

Answer 2

in the original equation, \[\alpha = \Delta \omega/ \Delta t\], is t = time? the radius is known but time is not.

Answer 3

t is time. What do you know?

Answer 4

force 84N @ angle of 20, radius 2.3, mass 75kg, use moment of inertia for solid cylinder I=1/2MR^2

Answer 5

Okay. We do not have to use the above expressions.
First, solve for the component of the force tangential to the wheel. I'll assume the angle is made from the tangential of the wheel to the force and say \[F_t = 84\cos(20)\]If the angle is made between the force and line that passes through the point where the force is applied and the center of the wheel it will be sin(20).

We know from translational Newtonian physics that \[F = m*a\]In rotational physics this can be expressed as \[F = I*\alpha\]where F is F_t.

Answer 6

Oops. It should be expressed as for rotational physics. \[\tau = I *\omega \]So since the force is known, we can write \[F_t*r = I*\omega\]

Answer 7

Dang it. \[\tau = I * \alpha \rightarrow F_t*r = I*\alpha\]Sorry about that. I don't know why I put omega.

Answer 8

Let me know what you get and I'll double check your answer.

Answer 9

using torque = I alpha

only frictional torque will be there

so if we know the coefficient of friction and the mass of ball,

then we can calculate alpha