Determine the angle between the inclined plane and the horizontal

[Gtseviper]

A hollow, thin-walled cylinder and a solid sphere start from rest and roll without slipping down an inclined plane of length 5 m. The cylinder arrives at the bottom of the plane 2.6 s after the sphere. Determine the angle between the inclined plane and the horizontal.

x=Vo + 1/2 at^2 and I got 3.97s for t
I=1/2 M(R^2 + R^2 I=2/5 MR^2
I got 5.20 for phida

The sun's radius is 6.96 108 m, and it rotates with a period of 25.3 days. Estimate the new period of rotation of the sun if it collapses with no loss of mass to become a neutron star of radius 5.3 km.

T2/T1 = R^2/R^2 and I got 1.47 x 10^-9 days m for T2 and then I converted it and get 1.27 x 10^-4ms
Figure 10-45 shows a hollow cylindrical tube of mass M = 0.8 kg and length L = 1.9 m. Inside the cylinder are two masses m = 0.4 kg, separated a distance = 0.6 m and tied to a central post by a thin string. The system can rotate about a vertical axis through the center of the cylinder. The system rotates at such that the tension in the string is 108 N just before it breaks.

M xWo^2 x r=T to get 30 rad/s for Wo
I initial=ML^2/10 + mr^2 +mr^2 and I got 1.01 kg m^2
I final=ML^2/10 +Mr^2 +Mr^2 and I got 2.96 kg m^2
Io Wo= I(final) W(final) and I got 10.25 rad/s for W(final)

 

[HallsofIvy]

x=Vo + 1/2 at^2 and I got 3.97s for t
I=1/2 M(R^2 + R^2 I=2/5 MR^2

For which? The cylinder or the sphere?

I got 5.20 for phida

I have no idea if this is right or wrong. You didn't say what
"phida" means.

The sun's radius is 6.96 108 m, and it rotates with a period of 25.3 days. Estimate the new period of rotation of the sun if it collapses with no loss of mass to become a neutron star of radius 5.3 km.

T2/T1 = R^2/R^2 and I got 1.47 x 10^-9 days m for T2 and then I converted it and get 1.27 x 10^-4ms

Okay, you used "conservation of angular momentum". Looks good.

Figure 10-45 shows a hollow cylindrical tube of mass M = 0.8 kg and length L = 1.9 m. Inside the cylinder are two masses m = 0.4 kg, separated a distance = 0.6 m and tied to a central post by a thin string. The system can rotate about a vertical axis through the center of the cylinder. The system rotates at such that the tension in the string is 108 N just before it breaks.

What is the question? In any case, you don't say what the radius of the cylinder is.

 

[M98Ranger]

The angle between the inclined plane and the horizontal can be determined using the given information. We know that the cylinder and sphere start from rest and roll without slipping, and the cylinder arrives at the bottom of the plane 2.6 seconds after the sphere. Using the equation x = Vo + 1/2 at^2, we can calculate the time it takes for the cylinder to reach the bottom of the plane, which is 3.97 seconds.

Next, we can use the moment of inertia formula I = 1/2MR^2 + 2/5MR^2 to calculate the moment of inertia for both the cylinder and sphere. Plugging in the given masses and radii, we get a moment of inertia of 5.20 for the cylinder.

Using the equation T2/T1 = R2/R1, we can calculate the new period of rotation for the sun if it were to collapse into a neutron star with a radius of 5.3 km. This gives us a new period of rotation of approximately 1.27 x 10^-4 milliseconds.

Lastly, for the system of the hollow cylindrical tube and two masses, we are given that the tension in the string is 108 N just before it breaks and the system is rotating at an angular velocity of 30 rad/s. Using the equation M x Wo^2 x r = T, we can calculate the moment of inertia for the system to be 1.01 kg m^2. Once the string breaks, the masses will move further away from the center, increasing the moment of inertia to 2.96 kg m^2. Using the equation Io x Wo = I(final) x W(final), we can solve for the final angular velocity, which comes out to be 10.25 rad/s.