Consdar tha depicted conical pendulum: amas on tha and at a strng of largth L, w

Question

Consdar tha depicted conical pendulum: amas on tha and at a strng of largth L, which is hxed to tha celing. Given tha prapar push, th pandulum can swing with an angua alnty in a cincla at an angla a with raspect to the vertical, maintairing the samc haight thrcughout is motion. Diffcrent positions of tha mass ara incicated by North, Wast, South, Enst (N, W 5, E Submt Answer y-may ubmt Anwer Tes 010 what i tha tensin on the cabla in terms of the argle a? ro) m*g/cos(a are correLL recaipt no. is 167-2628s Ti what i: thenguar velocity zquared in terms of thengle c? Submt Answer Tries 110 Pravicu8 Tes if the mas iz 5.8 kg, the angle 22.5 degrees, and the length of the cable 2.4 meters, what is the linear speed of the ball 3.12m uhmt Answar Trias 110 PravicR Tes

Explanation / Answer

radius of the circular path=R=L*sin(alpha)

let tension in the string be T.

centripetal force=mass*angular speed^2*radius=m*w^2*L*sin(alpha)

forces in x axis=0

force in y axis=T*sin(alpha)-m*w^2*L*sin(alpha)

force in z axis=T*cos(alpha)-m*g

balancing forces along z axis:

T*cos(alpha)-m*g=0

==>T=m*g/cos(alpha)

balancing forces along y axis:

T*sin(alpha)-m*w^2*L*sin(alpha)=0

==>w^2=T/(m*L)

=m*g/(cos(alpha)*m*L)

=g/(L*cos(alpha)

given mass=m=5.8 kg

alpha=22.5 degrees

L=2.4 m

then w=sqrt(g/(L*cos(alpha))

=2.1023 rad/s

then linear speed=w*R

=w*L*sin(alpha)

=1.931 m/s

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