# Calculating the torque requirements of the motor for a specific case

I want to use a DC 12/24V motor to rotate an assembly of glass tubes(image attached) through a hollow shaft of 25mm OD and 22mm ID, the shaft is supported on bearings on both the sides(assume minimum friction), two 3mm thick circular plates hold the four tubes together at both the end. The shaft is attached to the center of the disk.

Data:

• Glass tube: 58 mm OD, 45 mm ID 1.8 m long
• Disk: 180 mm OD, 3 mm thick attached at 25 mm from both the edges of the glass tube.

Even a general procedure/idea of calculating torque requirement is appreciated

• So how much does the whole rotating assembly weigh? How fast do you need it to spin continuously, and how quickly do you need it to reach that speed? Motor specification is linked to both peak torque (while accelerating) and steady-state speed. – Jonathan R Swift Jun 14 '18 at 13:00
• @JonathanRSwift, The whole assembly weighs 23kg, I need it to reach a 100RPM in 2.5 sec while starting and stay steady. – abhijeet savant Jun 15 '18 at 4:24

The torque required during the acceleration phase can be calculated using the formula:

$$T=Ia$$

where: $T = \text{Torque (Nm)}\\ I = \text{moment of inertia (kg.m²)}\\ a = \text{angular acceleration (rad/s²)}$

The moment of inertia of a solid disc is $I=\frac{1}{2}MR^2$, where $M=\text{mass (kg)}$, and $R=\text{radius (m)}$.

In this case, the moment of inertia will be lower than that of a solid disk, since the arrangement of the glass tubes concentrate the mass closer to the axis. Therefore, it is a safe estimate to use for calculation, and will ensure that the selected motor has sufficient power. Using the numbers provided in your question, the radius of the disc is found to be $0.09\text{ m}$

$$I=\frac{1}{2}(23)(0.09)^2=0.0932\text{ (kg.m²)}$$

Using the numbers that you provided in your comment, the angular acceleration can be calculated by converting the steady-state speed $100\text{RPM}=10.47\text{rad/s}$, and dividing by the 'spin up time' in seconds

$$a=\frac{10.47}{2.5}=4.19\text{ (rad/s²)}$$

Combining these two calculated values in the original equation gives the peak torque during acceleration:

$$T=(0.0932)(4.19)=0.39\text{ Nm}$$

It's important to note, however, that the torque generated by any motor is not constant in relation to angular velocity.

If you specify a motor which is advertised as being capable of sustaining $0.4\text{ Nm}$ at $100\text{ RPM}$, then you should be OK, since the available torque will be higher at lower speeds. This corresponds to a power of $\approx 4.2\text{ W}$, which is certainly achievable with a small DC motor.

If you want to limit the spin-up-speed to take $2.5\text{ s}$ in any case, then this should be done using e.g. PWM control of the motor.

• Thank you for your help @Jonathan R Swift, this was really helpful – abhijeet savant Jun 16 '18 at 9:03