Saturday

July 26, 2014

July 26, 2014

Posted by **rufy** on Friday, April 9, 2010 at 3:43pm.

What is the magnitude of the frictional force on the sphere?

--The other person did this for me but it's still saying it's wrong.

The speed acquired at the bottom is related to the height of the incline,

H = 3.4 sin 29 = 1.648 m

For a uniform-density sphere that is not slipping, conservation of energy requires that

(1/2)M V^2 + (1/2)(2/5)V^2 = M g H

V = sqrt(10/7)gH = 4.80 m/s

The acceleration rate (a) of the sphere is such that

V = sqrt(2 a X)

a = V^2/2X = 3.4 m/s^2

The angular acceleration rate is

alpha = a/R = 12.14 radian/s^2

The friction force can now be obtained from the equation relating angular acceleration to torque. The friction force F provides the torque needed to make it spin as it rolls dwn the plank.

F*R = I*alpha = (2/5)MR^2*alpha

F = (2/5)MR*alpha

= (0.4)*3.7 kg*0.18 m*12.14 s^-2

= 3.23 N

still saying it's wrong

|f| = N

3.23 NO

HELP: The frictional force provides the torque needed to give an angular acceleration. Therefore, first find the angular acceleration, then apply the rotational equivalent of Newton's 2nd Law (torque = I*a).

HELP: Since the sphere rolls without slipping, the angular acceleration is related to the translational acceleration, which can be found from kinematic relations.

- Physics typo -
**Damon**, Friday, April 9, 2010 at 3:47pm(1/2)M V^2 + (1/2)(2/5)V^2 = M g H

well that should be

(1/2)M V^2 + (1/2)(2/5)MV^2 = M g H

I think

- Physics -
**Damon**, Friday, April 9, 2010 at 3:52pmBut that is just a typo, the next line is good.

- Physics -
**Damon**, Friday, April 9, 2010 at 3:59pmThe angular acceleration rate is

alpha = a/R = 12.14 radian/s^2

agree so far

- found disagreement -
**Damon**, Friday, April 9, 2010 at 4:04pmF*R = I*alpha = (2/5)MR^2*alpha

F = (2/5)MR^2*alpha

= (0.4)*3.7 kg* .28^2 m^2 *12.14 s^-2

= 1.41 N

- Wow , another typo -
**Damon**, Friday, April 9, 2010 at 4:08pmF*R = I*alpha = (2/5)MR^2*alpha

F = (2/5)M*R*alpha (divide both sides by R)

= (0.4)*3.7 kg* .28m *12.14 s^-2

= 5.03 N

**Related Questions**

physics - A solid sphere of uniform density starts from rest and rolls without ...

Physics - A solid sphere of uniform density starts from rest and rolls without ...

Physics - A solid sphere of uniform density starts from rest and rolls without ...

physics - solid sphere of uniform density starts from rest and rolls without ...

Physics 121 - A solid sphere of uniform density starts from rest and rolls ...

physics - A solid sphere of radius 23 cm is positioned at the top of an incline ...

Physics - A uniform solid sphere rolls down an incline without slipping. If the ...

physics - A solid uniform sphere is released from the top of an inclined plane ....

physics - A 3.0-kg sphere with a radius of 6.0 cm rolls from rest without ...

Physics - A uniform solid sphere of mass 4kg and diameter 20cm initally at rest...