A block of mass 3.00kg slides on a friction less track. The track is shaped into

Question

A block of mass 3.00kg slides on a friction less track. The track is shaped into a vertical, circular loop as shown. The radius of the loop, r, is 1.021 m. At the TOP of the loop the speed of the block, V| is 4.00m/s; at the SIDE of the loop, midway between the TOP and BOTTOM the block's speed, v_2, is 6.00m/s. In the boxes provided, complete the free-body diagrams of the block at the following positions: 1) at the TOP of the loop and 2) at the SIDE of the loop. Represent all forces with labeled arrows. Indicate the direction of instantaneous acceleration with a squiggly arrow Calculate a_c2, the centripetal acceleration of the block at position #2, the SIDE of the loop. Calculate a_a, the tangential acceleration of the block at position #2, the SIDE of the loop. Calculate | a |, the magnitude of the total acceleration of the block at position #2, the SIDE of the loop. Calculate n_1, the normal force of the track on the block at position #1, the TOP of the loop. What is the speed of the block at position #3, the BOTTOM of the loop? (Use Conservation of Energy)

Explanation / Answer

a)

b) a_c2 = v2^2/r

= 6^2/(2*1.021)

= 17.6 m/s^2

c) a_t2 = g = 9.8 m/s^2

d) a_total = sqrt(a_c2^2 + a_t2^2)

= sqrt(17.6^2 + 9.8^2)

= 20.1 m/s^2

e) Apply, Fnet = N + m*g

m*a_rad = N + m*g

==> N = m*g - m*a_rad

= m*g - m*v1^2/r

= 3*9.8 - 3*4^2/1.021

= -17.6 N

|N| = 17.6 N

f) Apply conservation of energy

0.5*m*v1^2 + m*g*(2*r) = 0.5*m*v3^2

==> v3 = sqrt(v1^2 + 4*m*g*r)

= sqrt(4^2 + 4*3*9.8*1.021)

= 11.7 m/s

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