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Circular Motion and Newton's Seconds Law (Calculus)

  • Thread starter Johnny0290
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  • #1

Homework Statement



Consider a bead of mass m that is free to move around a horizontal, circular ring of wire (the wire passes through a hole in the bead). You may neglect gravity in this problem (assume the experiment is being done in space, far away from anything else). The radius of the ring of wire is r. The bead is given an initial speed v_0 and it slides with a coefficient of friction mu_k. In the subsequent steps we will investigate the motion at later times. You should begin by drawing a free-body diagram at some instant of time. Note that there will be a radial acceleration, a_R, and a tangential acceleration, a_T, in this problem.

1.1 Write Newton's 2nd law for the radial and tangential directions.

1.2 Combine the above equations to write a differential equation for dv/dt, where v is the speed at time t.

1.3 Solve the above differential equation to determine v(t). The solution has the form v = c1/(1+c2*t) - find c1 and c2. Hint: if v_0 = 3 m/s, mu_k = 0.1, r = 10 cm, and t = 3 s, v(3) = 0.3 m/s.


Homework Equations


F=ma
a_r=v^2/r
a_total=dv/dt
a_total^2=a_t^2 + a_r^2


The Attempt at a Solution


1.1
Radial Acceleration
F=ma
N=ma_r
m(v^2/r)=ma_r
a_r=v^2/r

Tangental Acceleration
F=ma
F_friction=ma_t
mu_k*N=ma_t
mu_k*m(v^2/r)=ma_t
a_t=mu_k*(v^2/r)

1.2
a_total^2=a_t^2 + a_r^2

After plugging in a_t and a_r and simplifying I came up with...
a=(v^2/r)*sqrt(1+(mu_k^2))

1.3
a=dv/dt

dv/dt=(v^2/r)*sqrt(1+(mu_k^2))

Rearranging the equation...
dv/v^2=(dt/r)*sqrt(1+(mu_k^2))

Integrating both sides...
-1/v=(t/r)*sqrt(1+(mu_k^2))+C

Solving for v...
v=-1/((t/r)*sqrt(1+(mu_k^2))+C)

Then I used the information in the hint to solve for C. C=-1/3
So...
v=-1/((t/r)*sqrt(1+(mu_k^2))-(1/3))

When I tried to use the information in the hint I was unable to rearrange the equation into the format the question asked for.
It also doesn't give me the correct answer when I plug in the numbers from the hint.

Where did I go wrong? Any help is appreciated. Thank You!!!
 

Answers and Replies

  • #2
vela
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Homework Statement



Consider a bead of mass m that is free to move around a horizontal, circular ring of wire (the wire passes through a hole in the bead). You may neglect gravity in this problem (assume the experiment is being done in space, far away from anything else). The radius of the ring of wire is r. The bead is given an initial speed v_0 and it slides with a coefficient of friction mu_k. In the subsequent steps we will investigate the motion at later times. You should begin by drawing a free-body diagram at some instant of time. Note that there will be a radial acceleration, a_R, and a tangential acceleration, a_T, in this problem.

1.1 Write Newton's 2nd law for the radial and tangential directions.

1.2 Combine the above equations to write a differential equation for dv/dt, where v is the speed at time t.

1.3 Solve the above differential equation to determine v(t). The solution has the form v = c1/(1+c2*t) - find c1 and c2. Hint: if v_0 = 3 m/s, mu_k = 0.1, r = 10 cm, and t = 3 s, v(3) = 0.3 m/s.


Homework Equations


F=ma
a_r=v^2/r
a_total=dv/dt
a_total^2=a_t^2 + a_r^2


The Attempt at a Solution


1.1
Radial Acceleration
F=ma
N=ma_r
m(v^2/r)=ma_r
a_r=v^2/r

Tangental Acceleration
F=ma
F_friction=ma_t
mu_k*N=ma_t
mu_k*m(v^2/r)=ma_t
a_t=mu_k*(v^2/r)

1.2
a_total^2=a_t^2 + a_r^2

After plugging in a_t and a_r and simplifying I came up with...
a=(v^2/r)*sqrt(1+(mu_k^2))

1.3
a=dv/dt

dv/dt=(v^2/r)*sqrt(1+(mu_k^2))

Rearranging the equation...
dv/v^2=(dt/r)*sqrt(1+(mu_k^2))

Integrating both sides...
-1/v=(t/r)*sqrt(1+(mu_k^2))+C

Solving for v...
v=-1/((t/r)*sqrt(1+(mu_k^2))+C)

Then I used the information in the hint to solve for C. C=-1/3
So...
v=-1/((t/r)*sqrt(1+(mu_k^2))-(1/3))

When I tried to use the information in the hint I was unable to rearrange the equation into the format the question asked for.
It also doesn't give me the correct answer when I plug in the numbers from the hint.

Where did I go wrong? Any help is appreciated. Thank You!!!
Only the tangential component of the acceleration affects the speed. The normal acceleration only affects the direction of the bead.
 
  • #3
Wow thank you so much... so far it's working out.
 

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