Calculating Angular Speed for a Gravitational Field of 1.95g

AI Thread Summary
To calculate the angular speed needed to create an effective gravitational field of 1.95g, the biologist must consider both Earth's gravity and the artificial gravity produced by the rotating table. The relevant equation is a = w^2r, where 'a' is the effective acceleration. The initial calculation of angular speed using the formula sqrt(a/r) resulted in 6.72 rad/sec, but this was identified as incorrect. The correct approach requires incorporating Earth's gravitational acceleration into the total effective gravitational field. The problem emphasizes the importance of accurately combining gravitational forces to determine the necessary angular speed.
sona1177
Messages
171
Reaction score
1

Homework Statement


A biologist studying plant growth and wants to stimulate a gravitational field stronger than the Earth's. She places the plants on a horizontal rotating table in her laboratory (on Earth) at a distance of 42.3 cm from the axis of rotation. What angular speed will give the plants an effective gravitational field , geff, of magnitude 1.95 g? [Hint: Remember to account for Earth's gravitational field as well as the artificial gravity when finding the effective gravitational field.


Homework Equations


a=w^2r



The Attempt at a Solution


sq rt (a/r)=w
sq rt (1.95g/.423)=w=6.72 rad/sec
This is wrong according to my text. How do i do this problem?
 
Physics news on Phys.org
sona1177 said:

Homework Statement


A biologist studying plant growth and wants to stimulate a gravitational field stronger than the Earth's. She places the plants on a horizontal rotating table in her laboratory (on Earth) at a distance of 42.3 cm from the axis of rotation. What angular speed will give the plants an effective gravitational field , geff, of magnitude 1.95 g? [Hint: Remember to account for Earth's gravitational field as well as the artificial gravity when finding the effective gravitational field.


Homework Equations


a=w^2r

The Attempt at a Solution


sq rt (a/r)=w
sq rt (1.95g/.423)=w=6.72 rad/sec
This is wrong according to my text. How do i do this problem?


...
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top