Requesting help with centripetal acceleration homework

AI Thread Summary
The discussion revolves around a physics homework problem involving a hockey puck and a mass hanging from a string. Participants clarify that the gravitational force on the puck does not affect the tension in the string because the tabletop is smooth, eliminating friction. The normal force from the table counteracts the puck's weight, leading to the conclusion that gravity's influence is negligible in this scenario. Additionally, there is an emphasis on the importance of including units for all variables in physics problems. The original poster expresses gratitude for the clarifications, indicating improved understanding.
Myani
Messages
2
Reaction score
0

Homework Statement


A hockey puck of mass m = 80 g is attached to a string that passes through a hole in the center of a table, as shown in the figure below. The hockey puck moves in a circle of radius r = 1.10 m. Tied to the other end of the string, and hanging vertically beneath the table, is a mass M = 1.00 kg. Assuming the tabletop is perfectly smooth, what speed must the hockey puck have if the mass M is to remain at rest?

M puck = 0.08kg
M block = 1kg
Gravity = 9.8
r = 1.10

Homework Equations



F=ma
Ac = (v^2/r)

The Attempt at a Solution



Attached in photo, why the gravity force of puck not being included is confusing me alot, shouldn't all objects be affected by gravity?
 

Attachments

  • helpp.PNG
    helpp.PNG
    34.8 KB · Views: 377
Physics news on Phys.org
Myani said:
Attached in photo, why the gravity force of puck not being included is confusing me alot, shouldn't all objects be affected by gravity?
Do you feel gravitational force on the buck has any influence on the tension of the string? Are they not acting perpendicular to each other over the buck? Moreover, the table surface is smooth, right?
 
You'll need someone else to verify if this is correct or not as my physics know-how is a bit dated these days. However, I believe gravity comes into play in the form of friction against the surface. Since we're assuming the tabletop is perfectly smooth, gravity on the puck is no longer a factor because there is no friction. Gravity's pull in this case is counteracted by the table.
 
Myani said:

Homework Statement


A hockey puck of mass m = 80 g is attached to a string that passes through a hole in the center of a table, as shown in the figure below. The hockey puck moves in a circle of radius r = 1.10 m. Tied to the other end of the string, and hanging vertically beneath the table, is a mass M = 1.00 kg. Assuming the tabletop is perfectly smooth, what speed must the hockey puck have if the mass M is to remain at rest?

M puck = 0.08kg
M block = 1kg
Gravity = 9.8
********** 9.8 what? **************
r = 1.10
********* 1.10 what ****************

Homework Equations



F=ma
Ac = (v^2/r)

The Attempt at a Solution



Attached in photo, why the gravity force of puck not being included is confusing me alot, shouldn't all objects be affected by gravity?

Units are important; do not state units for two of the data items and omit them for two others.
 
Myani said:
why the gravity force of puck not being included
Because the normal force from the table is equal and opposite to it, so they cancel. And as others have mentioned, there is no friction.
 
Thank you for all your responses, I am able to understand it now.

Also next time I will make sure to add all the units to my variables.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top