Recent content by jeff.reinecke

I figured it out I needed my KErot to be 2/5Iw^2 and realized that Vi = 4.08 not w =4.08 w = V/R Sorry I typed this on my phone

please help anybody

im still getting the wrong answer instead of having V2 = ((ωR)^2 -4gR)^(1/2) i now have V2 = (((ωR)2 -4gR)/R2)1/2 which ends up with a wrong answer

KErot = 0.5Iω^[2] is that not right since is do not have the mass of the ball i wouldn't not be able to solve equation because i would not be able to cancel out inertia

but wouldn't i need to add inertia to the equation?

i used 0.5mv^2 KE1 + GPE1 = KE2 + GPE2 GPE1 = 0 so then i got 0.5mV1^2 = m(0.5V2^2 + gh) h = 2R (masses cancel each other out) V1^2 - 2gh = V2^2 V2 = ((ωR)^2 -4gR)^(1/2) V2= ((4.08(m/s)*0.458m)^2 - 4*9.8(m/(s^2))*0.458m)

Homework Statement A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.08 m/s on a horizontal section of a track as shown in the figure below. It rolls around the inside of a vertical circular loop of radius r = 45.8 cm. given radius =0.458m ω= 4.08...

you helped so much thank you

i had no idea which way to put friction since is initially i did not know which direction acceleration would be. Even though obviously the larger mass would force the direction of acceleration to it.

with my FBD i got Ʃfm1= T1 -m1g=m1a Ʃfm3=T2 -m3g =m3a Ʃfm2=T2 +(-T1) +(-fk1) +fk2 + N + m(-g) =m2a T1 = m1a + m1g + fk2 T2 = m3a + m3g + fk2 fk1 = μm2g fk2 = μm2g from there i just substituted in the t1 and t2's equation since everything else...

Homework Statement there are three masses m2 is on a table top with coefficient of friction of 0.350 with two different masses on each side. They are connected by a mass-less string and friction-less pulley m1=4kg m2=1kg m3=2kg 1.]find the acceleration of all three masses...