Gears, flywheels, levers, falling blocks, gravity, inertia, inertia force, momentum, angular momentum, velocity, angular velocity, acceleration, angular acceleration, displacement, energy, force, torque, power, mechanical advantage, all the good stuff, although not sure this is the place for it I'm giving it a try here because the guys here are ace
It's mechanical in nature
please follow, I will write down the steps one by one.
2) the arc length of load side = 0.001745329252m
3) inertia = 10.8 because weight is 60kg and radius is 0.6m
4) then we actually make the arc length that the load side of lever makes the circumference of the gear that attaches to axle of the wheel! so its radius is the arc length over 2pi which equals 0.0002777777778m
5) tangential force on gear equals torque over radius so pick arbitrary acceleration 0.05 multiply by inertia 10.8 and the torque equals 0.54. then 0.54/0.0002777777778 equals force of 1944 newtons
point 5 continued - torque also equals F * r so; 1944 * length of the load side of lever (0.005) multiply and arrive at 9.72Nm of torque. divide this number by an arbitrary length of 0.3m so, 9.72/0.3 gives the force in newtons to achieve the force that needs to act on the end of the effort side of the lever which is 32.4N
6) find the distance that the effort side of lever make, so (s = r * rad) 0.3 * 0.3490658504 rad = s of 0.1047197551m, and actually make this length be the circumference of a driving gear on a compound gear (the smaller one is the driven gear) the radius of the driving gear is the circumference/2pi which turns out to be 0.1047197551/2pi = 0.01666666667m.
7)for clarification multiply force by radius of driving gear 32.4 by 0.01666666667 that equals 0.54Nm
8) make driven gear on compound 5 * smaller by dividing the radius by 5 0.01666666667/5 = 0.003333333334, and then force 5 * larger: 32.4 * 5 = 162N
9) multiply 162 * 0.003333333334 = 0.54
10) make second gear on compound (smaller gear) find its circumference 0.003333333334 * 2pi = 0.02094395103m and make lever two load side move this distance s = r * rad so s/rad equal radius or distance to pivot 0.02094495103/0.4390658504 = 0.06m
11) times the 0.06 by 60 because that was the original ratio of the first lever so keeping it all neat = 3.6m
12) next to find the force needed on the end of lever 2, divide the newtons on the load side by 60 because that's the ratio 162/60 = 2.7N
13) divide 2.7 by the gravitational constant 2.7/9.80665 = 0.2753233775kg
14) find height that the effort side of the 2nd lever when raised through 0.3490658504rad, so 3.6 * 0.3490658504rad = 1.256637061m
15) torque equals F *r. 2.7 * 3.6 = 9.72Nm as calculated earlier!!
16) energy; F * d 2.7 * 1.256637061 = 3.392920065Joules torque time theta 9.72 * 0.3490658504 = 3.392920066J mgh 0.2753233775 * 9.80665 * 1.256637061 = 3.392920065J to work out the equation for E = 1/2 I * w^2 we must find the angular velocity w2^2 = w1^2 + 2 alpha * theta (w1 is 0) = 2 * 0.05 * 2pi rad = 0.7926654595 take the root and it gives 0.8903176172rad/s so plug into energy equation E = 1/2 * 10.8 * 0.7926654595^2 which indeed equals 3.392920099J
so now for the question!!!
I want to change gear so that the flywheel remains accelerating at 0.05rad/sec^2 but so that the block takes 15 seconds to drop on the 19th time the block is dropped. the flywheel is free to freewheel and the block is manually lifted back to the top to it's starting posistion each time before it's dropped again. Thanks people!
It's mechanical in nature
please follow, I will write down the steps one by one.
- Starting with the flywheel on the left side of the drawing, the flywheel has a gear on the axle of it.
- The lever with load side length of 0.005m is moved through 20 degree (0.3940658504 rad)
and moves counter clockwise (the flywheel moves in the clockwise direction)
2) the arc length of load side = 0.001745329252m
3) inertia = 10.8 because weight is 60kg and radius is 0.6m
4) then we actually make the arc length that the load side of lever makes the circumference of the gear that attaches to axle of the wheel! so its radius is the arc length over 2pi which equals 0.0002777777778m
5) tangential force on gear equals torque over radius so pick arbitrary acceleration 0.05 multiply by inertia 10.8 and the torque equals 0.54. then 0.54/0.0002777777778 equals force of 1944 newtons
point 5 continued - torque also equals F * r so; 1944 * length of the load side of lever (0.005) multiply and arrive at 9.72Nm of torque. divide this number by an arbitrary length of 0.3m so, 9.72/0.3 gives the force in newtons to achieve the force that needs to act on the end of the effort side of the lever which is 32.4N
6) find the distance that the effort side of lever make, so (s = r * rad) 0.3 * 0.3490658504 rad = s of 0.1047197551m, and actually make this length be the circumference of a driving gear on a compound gear (the smaller one is the driven gear) the radius of the driving gear is the circumference/2pi which turns out to be 0.1047197551/2pi = 0.01666666667m.
7)for clarification multiply force by radius of driving gear 32.4 by 0.01666666667 that equals 0.54Nm
8) make driven gear on compound 5 * smaller by dividing the radius by 5 0.01666666667/5 = 0.003333333334, and then force 5 * larger: 32.4 * 5 = 162N
9) multiply 162 * 0.003333333334 = 0.54
10) make second gear on compound (smaller gear) find its circumference 0.003333333334 * 2pi = 0.02094395103m and make lever two load side move this distance s = r * rad so s/rad equal radius or distance to pivot 0.02094495103/0.4390658504 = 0.06m
11) times the 0.06 by 60 because that was the original ratio of the first lever so keeping it all neat = 3.6m
12) next to find the force needed on the end of lever 2, divide the newtons on the load side by 60 because that's the ratio 162/60 = 2.7N
13) divide 2.7 by the gravitational constant 2.7/9.80665 = 0.2753233775kg
14) find height that the effort side of the 2nd lever when raised through 0.3490658504rad, so 3.6 * 0.3490658504rad = 1.256637061m
15) torque equals F *r. 2.7 * 3.6 = 9.72Nm as calculated earlier!!
16) energy; F * d 2.7 * 1.256637061 = 3.392920065Joules torque time theta 9.72 * 0.3490658504 = 3.392920066J mgh 0.2753233775 * 9.80665 * 1.256637061 = 3.392920065J to work out the equation for E = 1/2 I * w^2 we must find the angular velocity w2^2 = w1^2 + 2 alpha * theta (w1 is 0) = 2 * 0.05 * 2pi rad = 0.7926654595 take the root and it gives 0.8903176172rad/s so plug into energy equation E = 1/2 * 10.8 * 0.7926654595^2 which indeed equals 3.392920099J
- Initial Setup:
- A lever with a load side length of 0.005 meters is moved to an angle of 20 degrees (0.3940658504 radians).
- This results in an arc length of 0.001745329252 meters.
- Inertia Calculation:
- The inertia is determined to be 10.8 based on a weight of 60 kg and a radius of 0.6 meters.
- Gear and Torque Calculations:
- The arc length of the lever is used to create the circumference of a gear attached to the wheel axle.
- The torque required to move the gear is calculated based on an arbitrary acceleration, resulting in a force of 1944 newtons.
- Torque is also calculated using the force exerted on the lever side, resulting in 9.72 Nm.
- The force needed on the effort side of the lever is calculated to be 32.4 N.
- Compound Gear System:
- The height of the effort side of the lever is found to be 0.1047197551 meters, which is used as the circumference of a driving gear.
- The radius of the driving gear is calculated to be 0.01666666667 meters.
- The driven gear is made smaller by dividing its radius by 5, resulting in a radius of 0.003333333334 meters.
- The force on the driven gear is calculated to be 162 N, and the torque is confirmed to be 0.54 Nm.
- Second Compound Gear System:
- The circumference of the smaller gear in the compound system is calculated.
- The distance to the pivot point of the lever is determined based on the gear circumference.
- The force needed on the end of the second lever is calculated to be 2.7 N.
- Energy and Angular Velocity Calculations:
- Energy is calculated using different methods: force times distance, torque times angle, and potential energy.
- The equation for rotational kinetic energy, E = ½ * I *ω^2, is derived and confirmed.
-
ok good but at this point I want to point out that the end of the of the second lever is raised first and when it drops it raises the second lever so the wheel would spin clockwise -
Equation: Theta (in radians) = Initial angular velocity (w1) * time (t) + 1/2 * Angular acceleration * time^2
Given values: Initial angular velocity (w1) = 0 Angular acceleration = 0.05 Theta (angle for one full revolution) = 2pi
Solution: 2pi = 0 * t + 1/2 * 0.05 * t^2
0.05 * t^2 = 2pi * 2
t^2 = (2pi * 2) / 0.05
t^2 = (4pi) / 0.05
t^2 = (4 * 3.14159) / 0.05
t^2 ≈ 12.56636 / 0.05
t^2 ≈ 251.3272
t ≈ sqrt(251.3272)
t ≈ 15.85330919 seconds -
Here's a summary of the drops along with the corresponding time durations, angular velocities, and rotational kinetic energies:
Drop 1:- Time: 15.85330919 sec
- Angular velocity: 0.8903176172 rad/sec
- Rotational kinetic energy: 3.392920066 J
- Time: 6.034654264 sec
- Angular velocity: 1.1920533 rad/sec
- Rotational kinetic energy: 7.673351778 J
- Time: 4.789753763 sec
- Angular velocity: 1.431540988 rad/sec
- Rotational kinetic energy: 11.06627184 J
- Time: 4.096099625 sec
- Angular velocity: 1.636345969 rad/sec
- Rotational kinetic energy: 14.45919192 J
- Time: 3.637605426 sec
- Angular velocity: 1.81822624 rad/sec
- Rotational kinetic energy: 17.85211197 J
- Time: 3.305439609 sec
- Angular velocity: 1.98349822 rad/sec
- Rotational kinetic energy: 21.24503203 J
- Time: 3.05044627 sec
- Angular velocity: 2.136020534 rad/sec
- Rotational kinetic energy: 24.63795209 J
- Time: 2.846692596 sec
- Angular velocity: 2.278355164 rad/sec
- Rotational kinetic energy: 28.03087216 J
- Time: 2.679018985 sec
- Angular velocity: 2.412306113 rad/sec
- Rotational kinetic energy: 31.42379223 J
- Time: 2.537888269 sec
- Angular velocity: 2.539200526 rad/sec
- Rotational kinetic energy: 34.81671229 J
- Time: 2.1416958802 sec
- Angular velocity: 2.660048466 rad/sec
- Rotational kinetic energy: 38.20963235 J
- Time: 2.311826942 sec
- Angular velocity: 2.775639813 rad/sec
- Rotational kinetic energy: 41.60255241 J
- Time: 2.219326175 sec
- Angular velocity: 2.886606122 rad/sec
- Rotational kinetic energy: 44.99547247 J
- Time: 2.13711318 sec
- Angular velocity: 2.993461718 rad/sec
- Rotational kinetic energy: 48.38839255 J
- Time: 2.063411596 sec
- Angular velocity: 3.096632298 rad/sec
- Rotational kinetic energy: 51.78131057 J
- Time: 1.996846826 sec
- Angular velocity: 3.196474639 rad/sec
- Rotational kinetic energy: 55.17423064 J
- Time: 1.936336444 sec
- Angular velocity: 3.293291461 rad/sec
- Rotational kinetic energy: 58.5671507 J
- Time: 1.881014774 sec
- Angular velocity: 3.3873422 rad/sec
- Rotational kinetic energy: 61.96007076 J
so now for the question!!!
I want to change gear so that the flywheel remains accelerating at 0.05rad/sec^2 but so that the block takes 15 seconds to drop on the 19th time the block is dropped. the flywheel is free to freewheel and the block is manually lifted back to the top to it's starting posistion each time before it's dropped again. Thanks people!
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