Physics
posted by Sara .
A tire cruises up an embankment at 30° for 1 meter before launching across Piranha Lake. What is the velocity of the tire at the top of the embankment in m/s?
HELP: Inital KE of the tire is 1/2*M*V2 + 1/2*I*(V/R)2; both rotational and translational
Final KE = Initial KE  M*g*h where h = 1m * sin(30°).
Also: Initial Velocity is 2.8m/s Mass=10kg radius=.3m and I calculated the angualr momentum to be 4.2
It seems like the help part spells it all out but I still don't get the right answer.
This is what I tried to do:
KEfinal= 1/2(10)(2.8)^2 + 1/2 (10) (2.8)^2  (10)(9.8)(1 sin 30)
KEfinal= 29.4
29.4= 1/2mv^2
V= sqr(29.4*2/10)=2.42 but this is not the righ answer...I don't know what I am doing wrong

Physics 
bobpursley
Look at the second term in your KEfinal line. Is that supposed to be rotational energy? I don't see moment of inertia anywhere in that line.

Physics 
Sara
Yes, it is supposed to be rotational energy 1/2Iw^2
I=mr^2 (not sure if tire is considerd a hoop) and w=(v/r)2
I combined the two and the r's canceled out...is this correct? 
Physics 
bobpursley
Yes, a tire is a hoop. I= mr^2
rotatioinal energy= 1/2 I w^2=1/2 mr^2 (v/r)^2= 1/2 mv^2
You are right.
Now, at the top, you did calculate the final KE right, however, the tire is still rotating. Final KE has two parts, translational, and rotational. To calculate v, I think you assumed rotational was zero. 
Physics 
Sara
I just noticed in the previous problem it said to assume that the tires is a solid cylinder
thus Inertia must be I=1/2mr^2
So
(for 1/2Iw^2) I get: 1/2(1/2mr^2)(V/R)^2 )
KEfinal= 1/2mv^2 + 1/4mv^2 mgh
KEfinal= (1/2)(10)(2.8)^2 + (1/4)(10)(2.8)^2  (10)(9.8)(1sin30)
KEfinal= 39.2 + 19.6 49= 9.8
KEfinal=9.8
But what now...how do I solve for V? 
Physics 
Sara
I don't get the correct answer still
This is what I tried:
KEfinal=9.8
9.8= 1/2 mv^2 + 1/4mv^2
9.8/ (0.5*10) + (0.25* 10) = 2V^2
1.307=2V^2
1.307/2= V^2
sqr(1.307/2)=V= 0.808 still not correct though 
Physics 
Sara
I got it!
9.8= 1/2mv^2 + 1/4mv^2
9.8= (3/4)(mv^2)
[9.8*(4/3)]/m= V^2= 1.307
sqr(1.307)=V= 1.14 
Physics 
Brittany
9.8= 1/2mv^2 + 1/4mv^2
9.8= (3/4)(mv^2)
[9.8*(4/3)]/m= V^2= 1.307
sqr(1.307)=V= 1.14
I have a question.. Why is there 1/2 and 1/4 in the first question.. are they just constants?
Respond to this Question
Similar Questions

Physics
The warranty on a new tire says that an automobile can travel for a distance of 92,000km before the tire wears out. The radius of the tire is 0.33m. How many revolutions does the tire make before wearing out? 
physics help
A car travels at 80 km/h on a level road in the positive direction of an x axis. Each tire has a diameter of 66 cm. Relative to a woman riding in the car and in unitvector notation, what are the velocity at the (a) center, (b) top, … 
Physics
As you approach the I90 exit in your uncle's little red sports car convertible, you happen to notice a cat crossing the road. Not that you have any particular love of cats (especially this one, whose snooty disposition seems to beckon … 
physics
driving on a hot day causes tire pressure to rise. what is the pressure inside an automobile tire at 288, if the tire has a pressure of 30 psi at 318k? 
physics
A car travels at 107 km/h on a level road in the positive direction of an x axis. Each tire has a diameter of 74.0 cm. Relative to a woman riding in the car and in unitvector notation, what are the velocity at the (a) center, (b) … 
Physics
Suppose a large dump truck tire has a surface area of 26.3 m^2. If a force of 1.58×10^7 Newtons acts on the inner tire, what is the absolute pressure inside the tire? 
Physics
Suppose a large dump truck tire has a surface area of 26.3 m^2. If a force of 1.58×10^7 Newtons acts on the inner tire, what is the absolute pressure inside the tire? 
Physics
A car travels at 80 km/h on a level road in the positive direction of an x axis. Each tire has a diameter of 66 cm. Relative to a woman riding in the car and in unitvector notation, what are the velocity at the (a) center, (b) top, … 
Trigonomentry
A ladder 35 foot long is leaning against an embankment, inclined 62.5 degrees to the horizontal. if the bottom of the ladder is 10.2 feet from the embankment, what is the distance from the top of the ladder down to the embankment to … 
Trigonomentry
A ladder 35 foot long is leaning against an embankment, inclined 62.5 degrees to the horizontal. if the bottom of the ladder is 10.2 feet from the embankment, what is the distance from the top of the ladder down to the embankment to …