Physics Problem 2 Wheels rolling without slipping?

The two wheels shown below are connected by a conveyor belt which turns without slipping around both wheels. Wheel one has a radius of 2.00 meters while wheel two has a radius of 3.00 meters. Wheel one accelerates at 0.300 revolutions per second squared for 20.0 seconds from an initial angular speed of 1.00 radians per second.

-Determine the final angular speed of wheel one.

-Determine the angle wheel one turned through during the twenty second period.

-Determine the speed of the conveyor belt at the end of the twenty seconds.

-Determine the angular speed of wheel two at the end of the twenty seconds.

Here's a picture:

I understand the first two questions, but its the last two that throw me for a loop. (no pun intended)

2 Answers

  • 8 years ago
    Favorite Answer

    Throw me for a loop. HA! You're killin' me.

    Wheel 1:

    ω = ωo + αt = 1rad/s + 0.3rev/s² * 2πrad/rev * 20s = 38.7 rad/s after 20 s. ← (a)

    Then the linear speed of the belt is (equal to the tangential speed of the wheel, which is)

    v = ωr = 38.7rad/s * 2m = 77.4 m/s ← (c)

    Θ = Θo + ωo*t + ½αt² = 0 + 1rad/s * 20s + ½ * (0.3*2πrad/s) * (20s)² = 397 rads ← (b)

    Given the linear speed of the belt, then wheel 2's angular speed is

    ω = v / r = 77.4m/s / 3m = 258 rad/s ← (d)

    barring computational error.

    (EDIT: I misread the acceleration as 3rev/s² on first run-through.)

  • 4 years ago

    Kinetic power = a million/2mv^2 (0.5 mass circumstances velocity squared) fifty 5 = a million/2m(3.2^2) a hundred and ten = 10.24m m = 10.74kg Circumference of circle = 2(pi)r (r=radius in metres) 2(pi)(0.09) = 0.565m If it travels 3.2metres in step with 2d then in 5 seconds it travels 16m So sixteen/0.565 = 28.3 revolutions

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