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Posted by on Monday, October 26, 2009 at 1:45pm.

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 - , Monday, October 26, 2009 at 1:58pm

    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 - , Monday, October 26, 2009 at 2:03pm

    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 - , Monday, October 26, 2009 at 2:13pm

    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 - , Monday, October 26, 2009 at 2:20pm

    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 - , Monday, October 26, 2009 at 2:36pm

    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 - , Monday, October 26, 2009 at 3:10pm

    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 - , Tuesday, November 3, 2009 at 11:22pm

    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?

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