A uniform solid sphere rolls down an incline.
(a) What must be the incline angle if the linear acceleration of the center of the sphere is to have a magnitude of 0.50g?
To determine the incline angle required for the linear acceleration of the center of the sphere to be 0.50g, we can use the principles of rotational and translational motion.
First, let's calculate the linear acceleration of the center of the sphere. The linear acceleration of the center of mass of a rolling sphere is given by the formula:
a = (2/5) * (alpha * R)
Where:
a is the linear acceleration of the center of the sphere,
alpha is the angular acceleration of the sphere, and
R is the radius of the sphere.
In this case, we know that the linear acceleration a should be 0.50g, where g is the acceleration due to gravity (approximately 9.8 m/s^2). Therefore:
0.50g = (2/5) * (alpha * R)
Next, let's relate the angular acceleration alpha to the incline angle.
When a solid sphere rolls down an incline without slipping, the relationship between the net torque and the moment of inertia of the sphere is given by:
tau = I * alpha
Where:
tau is the net torque acting on the sphere,
I is the moment of inertia of the sphere, and
alpha is the angular acceleration of the sphere.
The net torque acting on the sphere can be calculated as the product of the force of gravitational acceleration and the distance from the center of mass to the point of contact with the incline (radius of the sphere):
tau = m * g * R
Where:
m is the mass of the sphere.
Now we can substitute the equations to eliminate the angular acceleration alpha:
m * g * R = I * (0.50g) * R
The moment of inertia of a solid sphere is given by:
I = (2/5) * m * R^2
Substituting this equation in, we get:
m * g * R = (2/5) * m * R^2 * (0.50g) * R
Simplifying:
1 = (2/5) * (0.50) * R
Now we can solve for the radius R:
1 = (1/5) * R
R = 5
Hence, the radius of the sphere is 5 units.
Now that we know the radius, we can determine the incline angle by using the properties of geometry. In a right triangle formed by the incline and the horizontal, the side opposite the angle is the height (h) and the side adjacent to the angle is the distance along the incline (d). The tangent of the angle is given by:
tan(angle) = h/d
In this case, the height of the triangle is equal to the radius of the sphere, which we found to be 5 units. Therefore:
tan(angle) = 5/d
To find the incline angle, we need to determine the distance along the incline (d). This distance can be found by using the formula for the arc length of a circle given the angle alpha and radius R:
s = alpha * R
In this case, we want the sphere to roll down the incline, so the arc length s should be equal to the distance along the incline d. Hence:
d = alpha * R
By substituting this equation, we get:
tan(angle) = 5 / (alpha * R)
Since alpha is equal to the angular acceleration divided by the radius R, we can further simplify the equation:
tan(angle) = 5 / (0.50g * R)
Finally, we solve for the incline angle (angle):
angle = arctan(5 / (0.50g * R))
Plugging in the values, we get:
angle = arctan(5 / (0.50 * 9.8 * 5))
angle ≈ arctan(0.102)
Using a calculator or trigonometric table, we can find that the approximate value of the angle is:
angle ≈ 5.84 degrees
Therefore, the incline angle required for the linear acceleration of the center of the sphere to have a magnitude of 0.50g is approximately 5.84 degrees.