Diameter is given but will need radius in the center (where arrow is) of the drum Times below are for the disc the disc
Posted: Thu Jun 09, 2022 4:32 pm
Diameter is given but will need radius in the center (where arrow is) of the drum
Times below are for the disc the disc and the ring and the ring for each of those it’s the 100 grams that makes it accelerate. The other mass that’s haggling on there is a frictional mass in both of the pulleys one the weight hits the ground the time stops.
The data below is of the diameter and mass of the ring and others as such which was already captured by the lab. The radius is needed for both of the ring and the disc not the diameter. Radius also needed from the spinning drum pictured not the disc since it is rotated on that point. (Important) Also using the correct units following them through each of the equations steps using them correctly.
Inertia Data:
Diameter of the Drum = 4.88 cm
Number of Turns = 9 (Revolutions)
Disc Data:Diameter of the Disc = 25.58 cmMass of the Disc = 4658 gDisc times: t1=13.82 s, t2 = 13.85 s, t3 = 13.82 s, t4 = 13.90 s
Ring Data:Outer Diameter of the Ring = 25.49 cmInner Diameter of the Ring = 22.47cmMass of the Ring = 4267 gRing times: t1= 17.96 s, t2 = 17.96 s, t3 = 17.86 s, t4 = 17.89s
Disc and Ring Data:Disc and Ring times: t1= 22.93 s, t2 = 22.93 s, t3 = 22.92 s, t4 = 22.94 s
Time when it goes down which is: The ring The disc on top 100 gram weight
BST NEED RADIUS IN THIS CENTER
STAMP MASS ON THE RING 4267 G THE RING
The Disc 4658
The disc
Rotational Inertia Results (m) Radius of the drum
Time for Mass to Fall. Trial 1 (s) Time for Mass to Fall. Trial 2 (s) Time for Mass to Fall. Trial 3 (s) Time for Mass to Fall. Trial 4 (s) Average Time (s) Angular Acceleration (rads/s²) Net Torque (Nm) Experimental Rotational Inertia (kg m²) Ring Disc Disc & Ring
Mass of Specimen (kg) Inner Radius of Specimen (m) Outer Radius of Specimen (m) Calculated Rotational Inertia (kg m²) Percent Difference Explain any differences: D
Finally, the applied torque is the product of this force by the radius (r) of the drum around, which the string is wound. That is: t=mgr. (6) Opposed to this is a torque due to friction, which can be overcome by hanging a small additional mass on the string. The angular acceleration of the turntable may be determined assuming the motion to be uniformly accelerated. The time required to make (N) revolutions as the string unwinds may be found from the relationship 0=2AN, (7) and since the initial angular velocity is zero: Combining both equation seven and eight and solving for (a) yields: 22NN (8) (9)
Times below are for the disc the disc and the ring and the ring for each of those it’s the 100 grams that makes it accelerate. The other mass that’s haggling on there is a frictional mass in both of the pulleys one the weight hits the ground the time stops.
The data below is of the diameter and mass of the ring and others as such which was already captured by the lab. The radius is needed for both of the ring and the disc not the diameter. Radius also needed from the spinning drum pictured not the disc since it is rotated on that point. (Important) Also using the correct units following them through each of the equations steps using them correctly.
Inertia Data:
Diameter of the Drum = 4.88 cm
Number of Turns = 9 (Revolutions)
Disc Data:Diameter of the Disc = 25.58 cmMass of the Disc = 4658 gDisc times: t1=13.82 s, t2 = 13.85 s, t3 = 13.82 s, t4 = 13.90 s
Ring Data:Outer Diameter of the Ring = 25.49 cmInner Diameter of the Ring = 22.47cmMass of the Ring = 4267 gRing times: t1= 17.96 s, t2 = 17.96 s, t3 = 17.86 s, t4 = 17.89s
Disc and Ring Data:Disc and Ring times: t1= 22.93 s, t2 = 22.93 s, t3 = 22.92 s, t4 = 22.94 s
Time when it goes down which is: The ring The disc on top 100 gram weight
BST NEED RADIUS IN THIS CENTER
STAMP MASS ON THE RING 4267 G THE RING
The Disc 4658
The disc
Rotational Inertia Results (m) Radius of the drum
Time for Mass to Fall. Trial 1 (s) Time for Mass to Fall. Trial 2 (s) Time for Mass to Fall. Trial 3 (s) Time for Mass to Fall. Trial 4 (s) Average Time (s) Angular Acceleration (rads/s²) Net Torque (Nm) Experimental Rotational Inertia (kg m²) Ring Disc Disc & Ring
Mass of Specimen (kg) Inner Radius of Specimen (m) Outer Radius of Specimen (m) Calculated Rotational Inertia (kg m²) Percent Difference Explain any differences: D
Finally, the applied torque is the product of this force by the radius (r) of the drum around, which the string is wound. That is: t=mgr. (6) Opposed to this is a torque due to friction, which can be overcome by hanging a small additional mass on the string. The angular acceleration of the turntable may be determined assuming the motion to be uniformly accelerated. The time required to make (N) revolutions as the string unwinds may be found from the relationship 0=2AN, (7) and since the initial angular velocity is zero: Combining both equation seven and eight and solving for (a) yields: 22NN (8) (9)