Question

3, (35%) A 2-kg block of negligible dimensions is supported by a string of length&-0.4 m and slides with a horizontal velocity 10.7 m/s on the outside surface of a smooth cone, whose side is at a angle θ-21° from the vertical axis, as shown. a. How much of the cord can be retracted, at a slow and steady rate, through the tip of the cone before the block leaves the surface of the cone? poin b. What is the horizontal velocity of the block if that much of the cord is retracted through the tip of the cone? 0.7230 m/s For part a, use principle of conservation of angular momentum First, some trigonometry: ,i-4, sín9-0.1433 m No angular impulse, so angular momentum is conserved: For velocity when block leaves cone, need forces and accelerations, so need FBD, and vertical and normal coordinates should do the job And want N = 0: mg and T sin θ cose In m- Combine to eliminate unneeded and unknown T: cos e Combine with conservation of angular momentum to eliminate unknown v: Solve for r2: sin Bv, 0.1388 m gtan θ Finally 1,- = 0.3873 m, and cord retracted E,--1,-0.01270 m sin 6 For part b, use expression for v2 already developed above, or use cons. of ang. mom. again: ,--/58 tan θ-0.7230 m /s, and v,--Ly,-0.7230 m/s

What is the value of L_0? Only this part confuses me. It's not used at all and then all of a sudden it just appears.

Answer 1

don't worry about that it's just a typing mistake

value of l₀ and l₁ is same

3, (35%) A 2-kg block of negligible dimensions is supported by a string of length&-0.4 m and slides with a horizontal velocity 10.7 m/s on the outside surface of a smooth cone, whose side is at a angle θ-21° from the vertical axis, as shown. a. How much of the cord can be retracted, at a slow and steady rate, through the tip of the cone before the block leaves the surface of the cone? poin b. What is the horizontal velocity of the block if that much of the cord is retracted through the tip of the cone? 0.7230 m/s For part a, use principle of conservation of angular momentum First, some trigonometry: ,İ-4, sin θ: 0.1433 m No angular impulse, so angular momentum is conserved: For velocity when block leaves cone, need forces and accelerations, so need FBD, and vertical and normal coordinates should do the job And want N 0: mg and T sin θ cose In m- Combine to eliminate unneeded and unknown T: cos e Combine with conservation of angular momentum to eliminate unknown v: T1 Solve for r: sin Bv, 0.1388 m gtan θ Finally 1,- = 0.3873 m, and cord retracted E,--1,-0.01270 m sin θ For part b, use expression for v2 already developed above, or use cons. of ang. mom. again: ,--/58 tan θ-0.7230 m /s, and v,--Ly,-0.7230 m/s

as per conservation of angular momentum

$r_{1}.m.v_{1}=r_{2}.m.v_{2}$

$r_{1}=l_{0}*sin\Theta$

you may check the answer you will get the answer

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