Recent content by PhysicsDud

Alright so let's see now, if there is only PE at first then the equation would be: PE = KE(block) + KE(Cylinder) mgy = 1/2 m_b v^2 + 1/2 I_c omega^2 Am I getting there?

KE = KE + PE 0.5 m(cylinder) v^2 = 0.5 m(block) v^2 + m(block) gh v = .82m/s a = .41 m/s^2 Is this correct?

i don't understand what s and u represent? Also, for mechanical energy I tried: M=m1+m2 I = 0.5 m2 r^2 omega = v/r 0.5Mv1^2 + 0.5Iomega1^2 + Mgh1 = 0.5Mv2^2 + 0.5Iomega2^2 + Mgh2 E initial = m2gh = 2.0kg x 9.80 x 0 = 0 E final = m1gh + 0.5 m1v^2 + 0.5 Iomega^2 = m1gh + 0.5m1v^2...

Thank You! Thanks so much for your help!

Ok hopefully I've got it this time I = 1/3 ML1^2 + 1/3 ML2^2 = 1/3 (.132)(.3)^2 + 1/3 (.308)(.7)^2 = 0.064 kg.m^2

Ok I'm in desperate need of some direction! I have the following question in which I must answer twice once using rotational dynamics and then again using Conservation of Mechanical Energy. Question: A massless string is wrapped around a solid cylinder as shown in the diagram at the right...

Is the correct answer 0.0132 kg.m^2 Using 0.3m for the Length?

Inertia through end of uniform rod That formula is I = 1/3ML^2 So the answer is 0.147 kg.m^2 ??

Confused Just wondering how you came up with 15.9rad/s??

My Work So I was a little confused but I calculated using the formula I = mr^2 for both sides of the stick and then added them: I = (0.132 kg)(0.3m)^2 + (0.308 kg)(0.7m)^2 = 0.1628 kg.m^2

My work 1 revolution = 360 degrees = 2 x pie rad = 6.28 rad 25 revolutions = (25 rev) (2 x pie rad/rev) = 50 x pie rad = 157 rad omega = 157 rad/ 5.0 s = 31.4 rad/s Therefore, the angular velocity of the wheel at the start of the 5.0 s interval was 31.4 rad/s And angular velocity at...

Wanted to check this question also: A meter stick of mass 0.44 kg rotates, in the horizontal plane, about a vertical axis passing through the 30 cm mark. What is the moment of inertia of the stick? (Treat it as a long uniform rod) Answer: I = 0.1628 kg.m^2 Correct??

Just wanted to check to ensure I have calculated the correct answers for the following question: A rotating wheel accelerates uniformly at 6.2 rad/s2, and completes 25 revolutions during a 5.0 s interval. (a) What is the angular velocity of the wheel at the start of the 5.0 s interval...

Thanks Thanks so much for the help!

I'm not sure how to go about doing this question. Any help would be great. A window cleaner of mass 95 kg places a 22-kg ladder against a frictionless wall, at an angle 65° with the horizontal. The ladder is 10 m long and rests on a wet floor with a coefficient of static friction equal to...