# Simplifying slope-angular acceleration vs torque

## Homework Statement

In a lab we conducted in class we graphed angular acceleration vs torque, and found a positive linear correlation-easy enough. However I am getting rather stuck in simplifying the units of slope in terms of Kg, m, and s... I am fairly sure this will give me that angular acceleration equals torque times inertia but I have trouble with simplifying the slope AND with even getting torque and angular acceleration simplified to kg m and s. I would really appreciate some help! thanks!!

## Homework Equations

Torque=NM
NM=Kg*a*m
kg*a*m=kg*(m/s^2)*m
=kg*m^2/s^2

at/r=m/s^2/r
m^2/s^2

## The Attempt at a Solution

(kg*m^2/s^2)/(m^2/s^2)
=kg

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vela
Staff Emeritus
Homework Helper

## Homework Statement

In a lab we conducted in class we graphed angular acceleration vs torque, and found a positive linear correlation-easy enough. However I am getting rather stuck in simplifying the units of slope in terms of Kg, m, and s... I am fairly sure this will give me that angular acceleration equals torque times inertia but I have trouble with simplifying the slope AND with even getting torque and angular acceleration simplified to kg m and s. I would really appreciate some help! thanks!!

## Homework Equations

Torque=NM
NM=Kg*a*m
kg*a*m=kg*(m/s^2)*m
=kg*m^2/s^2

at/r=m/s^2/r
m^2/s^2
Your mistake is here. You have
$$\left[\frac{a_t}{r}\right] = \frac{\text{m}/\text{s}^2}{\text{m}} = \cdots$$ You shouldn't get m2/s2.

## The Attempt at a Solution

(kg*m^2/s^2)/(m^2/s^2)
=kg