 #1
Gtseviper
A hollow, thinwalled 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 1045 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)
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 1045 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)
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