A toy consists of two identical spheres connected by a string with negligible mass. The toy is thrown at an angle above the horizontal such that the string remains taut and both sphere are revolving counterclockwise in a vertical plane around the center of the string.

when the toy was released, the center of the string was moving with an initial speed of 15m/s at a 60 degree angle above the horizontal. What is the speed of the center of the string at the instant when the string reaches the top of its trajectory?

At the top, the vertical component of the velocity is 0 m/s. Assuming negligible air resistance, the acceleration in the x direction is 0 m/s². So the speed of the center of the string is equal to the initial horizontal component of the velocity.

vₓ = 15 m/s cos 60° = 7.5 m/s

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priyankatutor priyankatutor · Answer 2 · 2022-01-17T07:25:51+01:00

At the instant when the string reaches the top of its trajectory, the speed of the center of the string is 7.5 m/s.

The verticle component will be 0 m/s at the top. So, the speed will depend upon Horizontal force.

What is horizontal Force?

The force along the surface is called the horizontal force. The force along the horizontal surface can be given as,

[tex]V_x = s \rm \ cos \theta[/tex]

Where,

[tex]V_x[/tex] - Horizontal component of velocity

[tex]s[/tex] - speed = 15 m/s

[tex]\theta[/tex] - angle - 60°

Put the values in the formula,

[tex]V_x = 15 \rm \ m/s \rm \ cos 60^o[/tex]

Since the value of the cos 60° = 1/2,

So

[tex]V_x = 7.5 \rm \ m/s[/tex]

Therefore, at the instant when the string reaches the top of its trajectory, the speed of the center of the string is 7.5 m/s.

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