Question

A planetary gear system has a band brake that holds the ring fixed. The sun is driven clockwise at 800 rpm with a torque of 16 N-m. The arm drives a machine. The gears have a module of m =2 mm, and the pressure angle of these gears is 20 degrees. The ring has 70 teeth and planet has 20 teeth.

a) What is circular pitch p?
b) What is the number of teeth in the Sun gear?
c) Sketch the forces applied to the arm and gears
d) What is the output torque?
e) What torque must be applied to ring to keep it stationary?

Answer 1

a) Check definition below

b) Number of teeth in the sun gear is 30

c) The sketch is in the attached file

d) The output torque is 11.85 Nm

e) The torque that must be applied to the ring to keep it stationary is 5.15 Nm

Step-by-step explanation:

a) Circular pitch is the ratio of the circumference of the pitch circle to the number of teeth contained in the gear.

It is given by the relationship,
`P_c = (\pi d)/(N)\\`

Where N is the number of teeth in the gear and d is the diameter of the pitch.

b) The relationship between the number of teeth in the sun gear, ring gear and planet gear can be given as:

`N_(R) = N_(s) + 2 N_p\\`

Number of teeth in ring gear,
`N_R = 70`

Number of teeth in the planet gear,
`N_p = 20`

`70 = N_(s) + 2 (20)\\N_s = 70 - 40\\N_s = 30`

Number of teeth in the sun gear is 30

c) The sketch of the forces applied to the arm and gear is attached as a file below

d)

The speed of rotation of the sun gear,
`\omega_s = 800 rpm`

torque on the sun gear,
`t_s = 16 N-m`

We can get the output torque of the planet gear by assuming an efficiency of 100%, where the input power will equal the output power.

`t_s \omega_s = t_p \omega_p\\t_p = (t_s \omega_s)/(\omega_p)`

To calculate the rotational speed of the planet,
`\omega_p`, first calculate the rotational speed of the arm,
`\omega_a` using the relation below:

`\omega_a = (\omega_s N_s + \omega_R N_R)/(N_R + N_s)`

The rotational speed of the ring gear is 0,
`\omega_R = 0`

`\omega_a = (\omega_s N_s )/(N_R + N_s)\\\omega_a = (800 * 30 )/(70 + 30)\\\omega_a = 240 rpm`

Then calculate the rotational speed of the planet from the relation below:

`(N_p)/(N_s) = (\omega_s - \omega_a)/(\omega_p - \omega_a) \\(20)/(30) = (800-240)/(\omega_p - 240)\\\omega_p = 1080 rpm`

The output torque is then calculated from the relation below:

`t_p = (t_s \omega_s)/(\omega_p)\\t_p = (16 * 80)/(1080)\\t_p = 11.85 N-m`

The torque output = 11.85 N-m

e)

For equilibrium, Summation of the torques applied on the gear is zero

`t_R + t_s + t_p = 0\\`

The torque applied on the planet is applied anti-clockwisely,
`t_p = -11.85 N-m`

`t_R + 16 - 11.85 = 0\\t_R = 5.15 N-m`