- #1

ChiralSuperfields

- 1,316

- 141

- Homework Statement
- I am trying to derive the equation that my textbook presents (5.28), however, I notice that they don't use the right hand rule for the torque so there is a slight change in the sign in my derivation

- Relevant Equations
- ##v_r = r\omega## for is a condition for rolling without slipping where ##v_r## is the speed of the COM of the object

For this,

I don't understand why they don't have a negative sign as the torque to the friction should be negative. To my understanding, I think the equation 5.27 should be ##I\frac{d \omega}{dt} = -F_{friction}R## from the right hand rule assuming out of the page is positive.

Noting that ##f_k = \mu_kmg## and integrating both sides, I get the equation of motion ##\frac{-Rmg \mu_kt}{I} = \omega(t)##

I also get ##v(t) = v_0 - u_kgt##

So setting the two equations equal to each other in the relation for rolling motion:

##v(t) = R \omega (t) ##

I get ##t_r = \frac{v_0}{\mu_kg(1 - \frac{mR^2}{I})}##. Could someone please explain to me who is wrong and why?

Many thanks!

I don't understand why they don't have a negative sign as the torque to the friction should be negative. To my understanding, I think the equation 5.27 should be ##I\frac{d \omega}{dt} = -F_{friction}R## from the right hand rule assuming out of the page is positive.

Noting that ##f_k = \mu_kmg## and integrating both sides, I get the equation of motion ##\frac{-Rmg \mu_kt}{I} = \omega(t)##

I also get ##v(t) = v_0 - u_kgt##

So setting the two equations equal to each other in the relation for rolling motion:

##v(t) = R \omega (t) ##

I get ##t_r = \frac{v_0}{\mu_kg(1 - \frac{mR^2}{I})}##. Could someone please explain to me who is wrong and why?

Many thanks!