# When calculating a car's acceleration, is it ok to substitute Torque?

• inv
In summary, when calculating a car's acceleration, it is acceptable to substitute torque into the equation a = F/m as long as certain conditions are met, such as the angle between the radius and force vectors being 90 degrees. However, this equation is only valid for particle-like motion and may not accurately represent planar motion.

#### inv

Homework Statement
" When calculating a car's acceleration, is it ok to substitute Torque into a= F/m "
Relevant Equations
a= F/m

T= Fr
1. When calculating a car's acceleration, is it ok to substitute Torque into a= F/m

a= F/m
T= Fr

F= T/r

where

a= acceleration,
F= force,
m= mass,
T= Torque,

a= T/rm ?

What type of torque are you referring to?

• etotheipi
Lnewqban said:
What type of torque are you referring to?
I think there is only one type of torque ; do you mean, where is this torque being measured?

• etotheipi
I agree with @haruspex, it is tricky to say without more context. If we measure torques about some fixed coordinate system, and the car we model as a particle, then if the resultant force on the car is ##\vec{F}## we can write down ##\vec{F} = m\vec{a}## (or with magnitudes, ##F=ma##) in addition to ##\vec{\tau} = \vec{r} \times \vec{F}##. The latter also reduces to ##\tau = rF\sin{\theta}## if we take magnitudes.

Now if the angle ##\theta## between the ##\vec{r}## vector and the ##\vec{F}## vector is 90 degrees, then ##\tau = rF##, like you say. So if all these conditions are satisfied, ##\tau = rma## is I believe valid.

The condition that ##\vec{r}## always be orthogonal to ##\vec{F}## is the most restrictive one here. It means that your equation is fine for something like circular motion, but generally incorrect for most planar motion.

• Lnewqban