Anonymous
Anonymous asked in Science & MathematicsPhysics · 6 months ago

A weight with mass 100 g is connected to a rotatable rob through two inelastic strings. Both strings are 50 cm. If you extend both strings..?

they create the angle 45 degrees against the rod. When the rob rotates the weight moves in a circular path.

What is the tension force in the lower string when the weight is rotating with 1 lap/second?

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  • 6 months ago

    Call the upper string 1 and the lower string 2.

    50cm = 0.5m, so radius of motion is: r = 0.5cos(45º) = 0.3536m

    ω = 1lap/s = 2π rad/s = 6.283 rad/s

    Centripetal force = mω²r = 0.1 * 6.283² * 0.3536 = 1.396N

    Centripetal force is the sum of the horizontal components of T₁ and T₂:

    T₁cos(45º) + T₂cos(45º) = 1.396

    Divide through by cos(45º):

    T₁ + T₂ = 1.974 (equation 1)

    Vertically the forces balance:

    Upwards force = total downwards force

    T₁sin(45º) = T₂sin(45º) + mg

    T₁ - T₂ = mg/sin(45º)

    . . . . . . = 0.1*9.8/sin(45º)

    . . . . . . .= 1.386 (equation 2)

    Subtract equation 2 from equation 1:

    2T₂ = 1.974 - 1.386

    T₂ = 0.294N

    You should probably round to 1 sig. fig. (giving 0.3N) or possibly 2 sig. figs (giving 0.29N).

    I used g = 9.8m/s². If you need a different value it will alter the final answer slightly.

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  • 6 months ago

    1 lap/second => f=1

    m=0.1 kg

    R=0.5 m

    F - centripetal force

    T - tension of bottom string

    T' - tension of top string

    ________________________________________________

    ω(angular frequency)=2πf since f is 1 ω=2π

    v(velocity)=Rω=2Rπ

    a(centripetal acceleration)=v^2/R=4Rπ^2

    F=ma=4Rmπ^2

    From geometry we can conclude

    Tsin45°+mg=T'sin45°

    Tcos45°+T'cos45°=F and since value of sin45° is the same as of the cos45° we can replace it, thus getting two equations with two unknonws

    Tsin45°+mg=T'sin45° (I.)

    Tsin45°+T'sin45°=F (II.)

    expressing T' from the first equation T'=T+mg/(sin45°)

    and substituting it into the second Tsin45°+Tsin45°+mgsin45°/(sin45°) = 4Rmπ^2

    2Tsin45°+mg = 4Rmπ^2

    2Tsin45° = 4Rmπ^2-mg

    sin45° is equal to √2/2 thus giving

    √2T = 4Rmπ^2-mg

    T=(4Rπ^2-g)m/√2

    substituting the values we get

    T=0.7 Newtons

    • Doctor6 months agoReport

      Used wrong input for R. R=0.5sin45°. The final value of T shoud be 0.292 Newtons as given by Steve4Physics.

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  • JOHN
    Lv 7
    6 months ago

    T = tension in lower string, T’ = tension in upper string.

    Since motion is wholly in a horizontal plane,

    T’ – T = 0.1g....(1)

    For motion in a horizontal circle

    T’cos45° + T cos45° = 0.1 x (2π)² = 0.4π²....(2)

    Substituting for T’ from (1) in (2)

    T + 0.1g + T = (0.4π²)/ cos45°

    T = [(0.4π²)/ cos45° - 0.1g]/2

    T = 2.3N.

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