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

a) a = 3.783 m/s^2

b) F_f = 5.045 N

Step-by-step 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

Answer 2

`a_(cm) = 9.64m/s^2`

`Ff=6.42N`

Step-by-step 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.