Solving Rolling Sphere on Sloped Plane: Mass, Friction, & Acceleration

[MAPgirl23]

A hollow, spherical shell with mass 1.50 kg rolls without slipping down a slope angled at 40.0 degrees.

a) Find the acceleration.
b) Find the friction force.
c) Find the minimum coefficient of friction needed to prevent slipping.

** When the sphere rolls, the rate of rotation and the linear velocity of the center are related. If the center is moving with speed v, and the contact point is not moving, the angular velocity, w (really the Greek letter omega), is v/R. The kinetic energy is the sum of translational energy of the center of mass and the rotational energy

KE = (1/2)mv^2 + (1/2)Iw^2

You can substitute for w and you will have the kinetic enegry expressed in terms of constants (m, I, R) and v^2. Then you can use the expression for I of a spherical shell to replace I and R in terms of m. You know that as the sphere rolls down the plane, it loses potentail energy and gains an equal amount of kinetic energy. You should be able to find a relationship between the distance rolled and v^2 that involves a trig function of the tilt angle. Then you can use the formula for the change in v^2

(v_f)^2 - (v_i)^2 = 2as

to find the aceleration. From the acceleration, you can find the net force causing the acceleration, which is the component of gravity parallel to the plane minus the friction.

I try this method but still get the wrong answers. Am I using the wrong values or wrong equations? What do I do?

 

[Dr.Brain]

You know that as the sphere rolls down the plane, it loses potentail energy and gains an equal amount of kinetic energy. You should be able to find a relationship between the distance rolled and v^2 that involves a trig function of the tilt angle. Then you can use the formula for the change in v^2

If the sphere starts from rest , ofcourse the potential energy it loses, it gains kinetic energy+ rotational energy

$mgh= \frac{1}{2} mv^2 + \frac{1}{2} Iw^2$


This will give you velocity at the end of the slope.Maybe you didnot consider the rotational energy?... did you?

(v_f)^2 - (v_i)^2 = 2as

Maybe now you can apply it?...is the height of the slope given?...If given! use the angle of the slope to find the length of the distance travelled!

Show your work..

 

[OlderDan]

A hollow, spherical shell with mass 1.50 kg rolls without slipping down a slope angled at 40.0 degrees.

a) Find the acceleration.
b) Find the friction force.
c) Find the minimum coefficient of friction needed to prevent slipping.

** When the sphere rolls, the rate of rotation and the linear velocity of the center are related. If the center is moving with speed v, and the contact point is not moving, the angular velocity, w (really the Greek letter omega), is v/R. The kinetic energy is the sum of translational energy of the center of mass and the rotational energy

KE = (1/2)mv^2 + (1/2)Iw^2

You can substitute for w and you will have the kinetic enegry expressed in terms of constants (m, I, R) and v^2. Then you can use the expression for I of a spherical shell to replace I and R in terms of m. You know that as the sphere rolls down the plane, it loses potentail energy and gains an equal amount of kinetic energy. You should be able to find a relationship between the distance rolled and v^2 that involves a trig function of the tilt angle. Then you can use the formula for the change in v^2

(v_f)^2 - (v_i)^2 = 2as

to find the aceleration. From the acceleration, you can find the net force causing the acceleration, which is the component of gravity parallel to the plane minus the friction.

I try this method but still get the wrong answers. Am I using the wrong values or wrong equations? What do I do?

Hmmmmmm I seem to have seen those words somewhere before . I don't want to say so "over there", but this is a better place to bring your questions. What are you using for I, and what expression do you wind up with for KE in terms of mass and velocity? What expression did you get for $$v^2$$ in terms of distance moved down the plane by considering the change in potential energy?

 

[MAPgirl23]

I = 2/3 * (1.50 kg) * r^2
what is r if the angle 40 degrees is given? I used cos(40)
how do I find the height?

 

[Dr.Brain]

what is r if the angle 40 degrees is given? I used cos(40)
how do I find the height?

r is in no way related to the angle of the slope..! ...

Either height of the slope/wedge or the length of the sloping surface should be given ... One of them is reqd.

 

[OlderDan]

I = 2/3 * (1.50 kg) * r^2
what is r if the angle 40 degrees is given? I used cos(40)
how do I find the height?

r is the radius of the sphere, which is not given. It is not needed bacause of the relationship between v, r and $\omega$. When you calculate the rotational kinetic energy, r divides out. You should be able to find the total kinetic energy in terms of m, v and a numerical constant.