Recent content by nbroyle1

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    Calc Centripetal Acceleration: Rotational Motion & Vinyl Record Player

    how does the circumference help me I am confused??
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    Calc Centripetal Acceleration: Rotational Motion & Vinyl Record Player

    So I found that the acceleration is 1.745rad/s^2. Then I plugged it into the equation for average acceleration to find the velocity and got 6.28m/s... is this correct??
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    Calc Centripetal Acceleration: Rotational Motion & Vinyl Record Player

    360 degrees. So then in one second I think it travels 100 degrees??
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    Calc Centripetal Acceleration: Rotational Motion & Vinyl Record Player

    im not sure... So its traveling 2∏ in 3.6s right??
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    Rotational Motion and speed of a disc

    I meant I added it to the right side of the equation oops
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    Calc Centripetal Acceleration: Rotational Motion & Vinyl Record Player

    In the days before compact discs and MP3 players (ancient history!), music was recorded in scratches in the surface of vinyl-coated disks called records. In a typical record player, the record rotated with a period of 3.6 s. Find the centripetal acceleration of a point on the edge of the record...
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    Rotational Motion and speed of a disc

    The reason I did that in part c was because I subtracted θo to the right side of the equation. I thought the initial angular displacement would be 1 since it is one radian.
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    Rotational Motion and speed of a disc

    A 36.5-cm diameter disk rotates with a constant angular acceleration of 2.00 rad/s2. It starts from rest at t = 0, and a line drawn from the center of the disk to a point P on the rim of the disk makes an angle of 57.3° with the positive x-axis at this time. (a) Find the angular speed of the...
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    A person pushes a 10.0 kg lawn mower at constant speed with a force of

    I did the equation V=Vo+at so after I plugged in the information and solved for acceleration it was 1/2. If I plug that into f=ma I get 5 but that isn't right.
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    A person pushes a 10.0 kg lawn mower at constant speed with a force of

    yes but since the person is pushing on the lawn mower you add the vertical component of his push + mg and that gives you the normal force.
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    A person pushes a 10.0 kg lawn mower at constant speed with a force of

    I tried that and got a=1/2. so when I plugged that into f=ma I got f=10(1/2) but the answer is incorrect. am I forgetting something or is my calculations just wrong??
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    A person pushes a 10.0 kg lawn mower at constant speed with a force of

    A person pushes a 10.0 kg lawn mower at constant speed with a force of 82.0 N directed along the handle, which is at an angle of θ = 45.0° to the horizontal (Fig. 4-45). (b) Calculate the horizontal retarding force on the mower (c) Calculate the normal force exerted vertically upward on the...
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    Velocity under constant acceleration

    Ok thanks yea I was just trying to solve for the wrong variable. That equation is a little new to me Ill have to remember that one.
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