Question

A hollow spherical shell with mass 2.00kg rolls without slipping down a slope that makes an angle of 40.0^\circ with the horizontal. Find the magnitude of the acceleration a_cm of the center of mass of the spherical shell. Find the magnitude of the frictional force acting on the spherical shell.

Answer 1

Answer:

$a_{cm} = 9.64m/s^2$

$Ff=6.42N$

Explanation:

The sum of torque on the sphere is:

$m*g*sin\theta*R=I*\alpha$

$m*g*sin\theta*R=2/3*m*R^2*\alpha$

$m*g*sin\theta*R=2/3*m*R*a_{cm}$

Solving for a:

$a_{cm}=9.64m/s^2$

Now, the sum of forces will be:

$m*g*sin\theta-Ff=m*a_{cm}$

Solving for Ff:

$Ff=m*g*sin\theta-m*a_{cm}$

Ff=-6.42N    The negative sing tells us that it actually points downwards.

Answer 2

Answer:

a) a = 3.783 m/s^2

b)  F_f = 5.045 N

Explanation:

Given:

- Mass of shell m = 2.0 kg

- Angle of slope Q = 40 degrees

- Moment of inertia of shell I = 2/3 *m*R^2

Find:

a) Find the magnitude of the acceleration a_cm of the center of mass of the spherical shell.

b) Find the magnitude of the frictional force acting on the spherical shell.

Solution:

- Draw a Free body diagram for the shell. We see that the gravitational force F_g acting parallel to the plane of the inclined surface makes the sphere to roll down. The frictional force  F_f between the inclined surface and the sphere gives the necessary torque for the sphere to roll down with out slipping. Under this conditions a sphere will roll down without slipping with some acceleration and the acceleration can be calculated from the equation of motion of the sphere:

                                      m*g*sin(Q) - F_f = m*a

- Where, The frictional force produces the torque and due to this torque the sphere gets an angular acceleration.

- Then we can write the equation for the rotational motion as:

                                      F_f*R = I*α

                                      F_f = I*α / R

- Using moment mass inertia of the shell we have:

                                      F_f = (2/3)*m*R^2*α/R

- Where the angular acceleration α is related to linear acceleration a with:

                                      α = a / R

- combing the two equations we will have friction force F_f as:

                                      F_f = (2/3)*m*R^2*a/R^2

                                      F_f = (2/3)*m*a

- Now evaluate the equation of motion:

                                      m*g*sin(Q) - (2/3)*m*a= m*a

- Simplify:

                                        (5/3)*a = g*sin(Q)

                                        a = (3/5)*g*sin(Q)

- Plug the values in:       a = (3/5)*9.81*sin(40)

                                        a = 3.783 m/s^2

- Now compute the Frictional force F_f from the expression derived above:

                                        F_f = (2/3)*m*a

- Plug values in:              F_f = (2/3)*2*3.783

                                        F_f = 5.045 N