Solving Rotational Mechanics: Minimum Friction & Kinetic Energy

In summary, LaTeX code ending with [/tex] instead of [\tex] results in a space being inserted between the text and the code.
  • #1
prasanna
45
0
Hey !
Please help me out with this problem.

A hollow sphere of mass m is released from top on an inclined plane of inclination [tex]\theta[\tex].
(a) What should be the minimum coefficient of friction between the sphere and the plane to prevent skidding?
I did this:

mgsin[tex]\theta[\tex] - f = ma ________ 1
(f is frictional force)

torque = I[tex]\alpha[\tex] = r X f
(symbols stand for their usual meanings)

r*f = [tex]\frac{2mr^2}{\frac{a}{r}}[\tex]

f = (2/3)ma _______2

Subst. in 1 ,
a= (3/5)gsin[tex]\theta[\tex] ______ 3

subst. both 2 and 3 in 1,

f = (2/5) mgsin[tex]\theta[\tex]
but,
[tex]\mu[\tex]mgcos[tex]\theta[\tex] = f = (2/5)mgsin[tex]\theta[\tex]

[tex]\mu[\tex] = (2/5) tan [tex]\theta[\tex]

this matches with the textbook answer.

(b) Find the Kinetic Energy of the ball as it moves down a length L on the incline if the friction coefficient is half the value calculated in part(a).
I did this :

[tex]\mu[\tex] = (1/5) tan [tex]\theta[\tex]

f = [tex]\mu[\tex]mgcos[tex]\theta[\tex]

putting this in 1(of part a)

a = (4/5) g sin[tex]\theta[\tex]

torque = I[tex]\alpha[\tex] = [tex]\frac{2mr^2}{3}[\tex]
[tex]\alpha[\tex] = [tex]\frac{3gsin\theta}{10r}[\tex]

KE = (1/2)mv^2 + (1/2)I[tex]\omega^2[\tex]

v^2 = 2aL = (8/5) gLsin[tex]\theta[\tex]

[tex]\omega^2[\tex] = 2[tex]\alpha\theta[\tex]

i found [tex]\omega^2[\tex] = (6gLsin[tex]\theta[\tex]) / (20[tex]\pi\r^2[\tex]

I get

KE = (4/5) mgLsin[tex]\theta[\tex] + mmgLsin[tex]\theta[\tex] / 2[tex]\pi[\tex]

KE = 0.831 * mgLsin[tex]\theta[\tex]

But the answer in the book is (7/8) mgLsin[tex]\theta[\tex]
which is 0.875 mgLsin[tex]\theta[\tex]


Where did I go wrong?
 
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  • #2
Hey !
Please help me out with this problem.

A hollow sphere of mass m is released from top on an inclined plane of inclination [tex]\theta[/tex].
(a) What should be the minimum coefficient of friction between the sphere and the plane to prevent skidding?
I did this:

mgsin[tex]\theta[/tex] - f = ma ________ 1
(f is frictional force)

torque = I[tex]\alpha[/tex] = r X f
(symbols stand for their usual meanings)

r*f = [tex]\frac{2mr^2}{\frac{a}{r}}[/tex]

f = (2/3)ma _______2

Subst. in 1 ,
a= (3/5)gsin[tex]\theta[/tex] ______ 3

subst. both 2 and 3 in 1,

f = (2/5) mgsin[tex]\theta[/tex]
but,
[tex]\mu[/tex]mgcos[tex]\theta[/tex] = f = (2/5)mgsin[tex]\theta[/tex]

[tex]\mu[/tex] = (2/5) tan [tex]\theta[/tex]

this matches with the textbook answer.

(b) Find the Kinetic Energy of the ball as it moves down a length L on the incline if the friction coefficient is half the value calculated in part(a).
I did this :

[tex]\mu[/tex] = (1/5) tan [tex]\theta[/tex]

f = [tex]\mu[/tex]mgcos[tex]\theta[/tex]

putting this in 1(of part a)

a = (4/5) g sin[tex]\theta[/tex]

torque = I[tex]\alpha[/tex] = [tex]\frac{2mr^2}{3}[/tex]
[tex]\alpha[/tex] = [tex]\frac{3gsin\theta}{10r}[/tex]

KE = (1/2)mv^2 + (1/2)I[tex]\omega^2[/tex]

v^2 = 2aL = (8/5) gLsin[tex]\theta[/tex]

[tex]\omega^2[/tex] = 2[tex]\alpha\theta[/tex]

i found [tex]\omega^2[/tex] = (6gLsin[tex]\theta[/tex]) / (20[tex]\pi\r^2[/tex]

I get

KE = (4/5) mgLsin[tex]\theta[/tex] + mmgLsin[tex]\theta[/tex] / 2[tex]\pi[/tex]

KE = 0.831 * mgLsin[tex]\theta[/tex]

But the answer in the book is (7/8) mgLsin[tex]\theta[/tex]
which is 0.875 mgLsin[tex]\theta[/tex]


Where did I go wrong?
Might help a bit ;)
 
  • #3
ponjavic said:
Might help a bit ;)

Thank a lot !
How did you do that?

What happened that my original thread did not show LaTEX ??
 
  • #4
Hey Prasanna, your name is like that of a cricketer.
Anyway, use [/tex] instead of [\tex] for ending LaTeX code.



spacetime
www.geocities.com/physics_all
 
  • #5
You used backslashes in the [/tex] instead of the forward slashes.

Backslashes for TeX characters, forward slashes for tags.

--J
 

1. What is rotational mechanics?

Rotational mechanics is a branch of physics that focuses on the study of rotation and its effect on the motion of objects. It involves principles such as torque, angular velocity, and angular momentum.

2. How do you calculate minimum friction in rotational mechanics?

To calculate minimum friction in rotational mechanics, you need to first find the normal force acting on the object and then use the coefficient of friction to determine the maximum friction force. If the applied force is less than the maximum friction force, then the minimum friction can be found by subtracting the applied force from the maximum friction force.

3. What is the role of kinetic energy in rotational mechanics?

Kinetic energy plays a crucial role in rotational mechanics as it represents the energy an object possesses due to its motion. In rotational motion, kinetic energy is dependent on both linear velocity and rotational velocity, and it is important in determining the total energy of a rotating object.

4. How does rotational motion differ from linear motion?

Rotational motion differs from linear motion in that it involves rotation around a fixed axis, while linear motion involves movement in a straight line. Additionally, rotational motion is affected by principles such as torque and angular momentum, while linear motion is affected by concepts such as force and momentum.

5. What are some real-world applications of rotational mechanics?

Rotational mechanics has many real-world applications, including the motion of objects such as wheels, gears, and pulleys. It also plays a role in the design and functioning of machines, vehicles, and sports equipment such as bicycles and footballs.

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