Similar to the situation shown in figure (a) below, a toy plane is being whirled

ID: 1533812 • Letter: S

Question

Similar to the situation shown in figure (a) below, a toy plane is being whirled at constant speed in a horizontal circle. The plane is attached to a string of length L. Figure (b) shows the relevant free-body diagram; figure (c) shows the coordinate system; and figure (d) shows the free-body diagram with the force of tension split into components. FT cos(6) FT sin(6) (a) Use g = 10 m/s, a speed for the plane of 1.60 m/s, and the angle = 34.0 degrees. Determine the length of the string (b) The mass of the plane is 0.223 kg. Calculate the tension in the string 2.64 (c) It may be a little surprising to you to note that the magnitude of the tension in the string is actually larger than the magnitude of the force of gravity acting on the plane. In which other situations is this also true? (Select all that apply.) The plane is at rest, hanging down from the string The string and the plane are set up like a pendulum, and we are interested in the tension in the string when the plane passes through the lowest point in its swing The string is used to suspend the plane from the ceiling of an elevator, and the elevator has an acceleration directed up none of the above

Explanation / Answer

part a:

let tension in the string be T N.

balancing the forces in vertical directon:

T*cos(theta)=m*g

==>T=m*g/cos(theta)...(1)

radius of the circle=L*sin(theta)

balancing forces in horizontal direction:

T*sin(theta)=m*v^2/r=m*v^2/(L*sin(theta))

==>L=m*v^2/(T*sin^2(theta))

=m*v^2*cos(theta)/(m*g*sin^2(theta))

=v^2*cos(theta)/(g*sin^2(theta))

using the values provided,

L=1.6^2*cos(34)/(10*sin^2(34))

=0.67872 m

part b:

T=m*g/cos(theta)

=0.223*10/cos(34)=2.6899 N

part c:
in second option, when the pendulum is passing through its lowest psoitino,

T=m*g+(m*v^2/r)

hence T > m*g

so second option is correct.

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Similar to the situation shown in figure (a) below, a toy plane is being whirled

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