How Far Back Must the Spring Be Pulled to Swing the Masses Around the Pivot?

  • Thread starter Thread starter Mercfh
  • Start date Start date
  • Tags Tags
    Rotational Spring
AI Thread Summary
To determine how far back the spring must be pulled to ensure the combined masses swing fully around the pivot, the minimum velocity required for the masses to complete the loop is calculated using the formula sqrt(5gR). The spring's launch distance can be determined using the equation d = Kx^2/(2mugm), where K is the spring constant, and µ is the coefficient of kinetic friction. The velocity after traveling 1 meter across the surface must be calculated, factoring in friction, to ensure it meets the minimum velocity for the loop. There is uncertainty about the correct equations to use, particularly regarding the inclusion of friction in the velocity calculations. The discussion emphasizes the importance of understanding energy transformations and gravitational potential energy in this context.
Mercfh
Messages
1
Reaction score
0

Homework Statement


A 0.2 kg mass is held against a spring with spring constant k=1000N/m. It is launched from the spring
and travels an additional 1 m across a surface with coefficient of kinetic friction µk=0.2. It then collides
and sticks to a 0.3 kg mass that is suspended from a 0.5 m long thread of negligible mass. How far back
must the spring be pulled (∆x) in order that the combined masses swing fully around the support pivot?

Homework Equations


Velocity Min to get around a loop= sqrt(5gR)
Spring Distance Launched= d = Kx^2/(2mugm)
Velocity of a Spring? = V^2=(k/m)*d^2
Distance = x = V^2/2a
Acceleration = m*g


The Attempt at a Solution


Well I wasn't sure if the Velocity equation was the "right" equation to find this. ideally I would "think" you would find the minimum speed that the spring would launch...but this doesn't put in mu for the velocity, so I am not sure how you calculate that. But once you find the velocity after 1m I figure you'd just check and see if that meets the minimum velocity to spin the object around the pivot.
I could be COMPLETELY wrong however
 
Physics news on Phys.org
i don't think you should use the escape velocity equation (??) to do this. hint: find the gravitational potential energy gained :D
 
Last edited:
Thread 'Variable mass system : water sprayed into a moving container'

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}...
View full post »

Similar threads

Question about springs and rotational/translational energy
Replies
4
Views
1K
Why can we move the spring with constant speed when we apply a force?
Replies
29
Views
2K
Force transmitted to a block in viscous fluid through a spring, cord, and pulley
Replies
6
Views
829
What Is the Frequency of Small Oscillations for a Pivoted Rod with Two Springs?
Replies
19
Views
2K
An error made in a kinematics question about a spring launcher?
Replies
16
Views
1K
How to solve a differential equation for a mass-spring oscillator?
Replies
2
Views
2K
Projectile Motion : Work and Energy
Replies
4
Views
1K
Help with this problem about a mass oscillating on a spring
Replies
1
Views
1K
Calculating spring constant of a spring loaded cannon
Replies
7
Views
3K
How to find the deceleration of a mass colliding on a spring?
Replies
5
Views
5K

Hot Threads

Recent Insights

Back
Top