Circular Motion and Newton's Seconds Law (Calculus)

In summary: Thank you for your time and help, it's greatly appreciated. In summary, the problem involves a bead of mass m moving on a horizontal, circular ring of wire with an initial speed v_0 and a coefficient of friction mu_k. After deriving the equations for the radial and tangential accelerations, the differential equation for dv/dt is found to be dv/dt=(v^2/r)*sqrt(1+(mu_k^2)). After solving this equation, the final solution for v(t) is v(t)=-1/((t/r)*sqrt(1+(mu_k^2))+C), where C is found to be -1/3 using the given information. However, only the tangential acceleration affects the speed
  • #1
Johnny0290
9
0

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!
 
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  • #2
Johnny0290 said:

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.
 

1. What is circular motion and how does it relate to Newton's Second Law?

Circular motion is the movement of an object along a circular path. According to Newton's Second Law, the net force acting on an object is equal to its mass multiplied by its acceleration. In circular motion, this acceleration is directed towards the center of the circle and is known as centripetal acceleration. Therefore, the net force acting on an object in circular motion is always directed towards the center of the circle.

2. How is calculus used in understanding circular motion and Newton's Second Law?

Calculus is used to analyze and measure the rate of change in circular motion. By taking derivatives and integrals of the position, velocity, and acceleration equations, we can calculate important quantities such as angular velocity and centripetal acceleration. These calculations help us better understand the relationship between circular motion and Newton's Second Law.

3. What is the difference between linear and circular motion in terms of Newton's Second Law?

The main difference between linear and circular motion is the direction of the acceleration. In linear motion, the acceleration is in the same direction as the net force, while in circular motion, the acceleration is directed towards the center of the circle. This means that the net force in circular motion is always changing direction, whereas in linear motion it remains constant.

4. What is the role of inertia in circular motion and Newton's Second Law?

Inertia is the tendency of an object to resist changes in its motion. In circular motion, inertia plays a role in determining the magnitude of the centripetal force needed to keep an object moving along a circular path. According to Newton's Second Law, the greater the mass of an object, the greater the force needed to accelerate it. Therefore, a larger mass will require a larger centripetal force to maintain circular motion.

5. How does circular motion and Newton's Second Law apply to real-life situations?

Circular motion and Newton's Second Law apply to many real-life situations, such as the motion of planets in orbit around the sun, the motion of a car around a curved track, and the motion of a roller coaster. Understanding these principles can also help in engineering and designing objects that require circular motion, such as centrifuges and amusement park rides.

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