A hollow, thin-walled cylinder and a solid sphere start from rest and roll without slipping down an inclined plane of length 5.0 m. The cylinder arrives at the bottom of the plane 2.7 s after the sphere. Determine the angle between the inclined plane and the horizontal.
To determine the angle between the inclined plane and the horizontal, we need to use the information given about the cylinder and the sphere.
Let's consider the motion of the cylinder first. Since it rolls without slipping, its linear acceleration down the incline can be related to its angular acceleration using the equation: a = α * r, where 'a' is the linear acceleration, 'α' is the angular acceleration, and 'r' is the radius of the cylinder.
Now, let's consider the motion of the sphere. Since it also rolls without slipping, its linear acceleration down the incline can be related to its angular acceleration using the same equation as the cylinder: a = α * r, where 'a' is the linear acceleration, 'α' is the angular acceleration, and 'r' is the radius of the sphere.
We can equate the linear accelerations of the cylinder and the sphere since they both roll without slipping. Therefore, a_sphere = a_cylinder.
The linear acceleration can be related to the angle of the incline using the equation: a = g * sin(θ), where 'g' is the acceleration due to gravity and θ is the angle between the inclined plane and the horizontal.
Let's denote the time it takes for the cylinder to reach the bottom as t_cylinder and for the sphere as t_sphere.
We know that the length of the incline is 5.0 m, so the linear distance traveled by both the cylinder and the sphere is the same. Therefore, we can write: a_cylinder * t_cylinder = a_sphere * t_sphere.
Plugging in the equations for linear acceleration and rearranging the equation, we have: g * sin(θ) * t_cylinder = g * sin(θ) * t_sphere.
Since the acceleration due to gravity 'g' and the sine of angle θ are the same for both objects, we can cancel them out, resulting in: t_cylinder = t_sphere.
From the given information, we know that the cylinder arrives at the bottom 2.7 seconds after the sphere: t_cylinder = t_sphere + 2.7.
Now we can solve the equation for t_sphere:
t_sphere = t_cylinder - 2.7.
Since t_cylinder = t_sphere, we can substitute that into the equation:
t_cylinder = t_cylinder - 2.7.
Solving for t_cylinder, we get: t_cylinder = 2.7 seconds.
Now that we have the time, we can use it to calculate the angle θ.
From the equation a = g * sin(θ), we rearrange it to: sin(θ) = a / g.
Using the linear acceleration of the cylinder, a_cylinder, we have: sin(θ) = a_cylinder / g.
Substituting in the values we know, we have: sin(θ) = (g * t_cylinder) / g.
Canceling out g, we get: sin(θ) = t_cylinder.
Now, we can find the angle θ by taking the inverse sine (arcsine) of both sides: θ = arcsin(t_cylinder).
Plugging in the value for t_cylinder, we have: θ = arcsin(2.7).
Using a calculator or trigonometric table, we can find the angle θ.
5m=1/2(5/7*(9.81*sin(theta))*((2.7*sqrt(1/2))^2)/((sqrt(5/7)-sqrt(1/2)))^2)
All you need to do is solve for sin(theta)
you do 5*((sqrt(5/7)-sqrt(1/2)))^2)*2*7/((5*9.81*((2.7*sqrt(1/2))^2))
sin(theta)=0.00746
then you do the inverse of sin(theta) to get your degree. degree should equal 0.428