A small block of mass m= 2.50 kgis released, starting at rest, from a heighthabo

Question

A small block of mass m= 2.50 kgis released, starting at rest, from a heighthabove the groundon a ramp inclined at 45.0degree (see figure below). The block reaches the bottom of the ramp andenters a loop-the-loop of radius R= 1.50 m. There is no friction between the block and thetrack. Treat the block as a point mass.

a)What is the speed of the block at position B if h= 9.00 m? (Notice: At position B, theblock is 2R above the ground.)

B) What is the magnitude of the normal force acting on the block when it is at positionB,the top of the loop, having started at the height h= 9.00 m? (Hint: Be sure to draw afree-body diagram for the mass at position B.)

c)Find the minimum starting height,hmin, for which the block will just make it through theloop without leaving the track at B. (Hint: When the blockjustmakes it through theloop, we set the normal force at B equal to zero, the critical condition, allowing us to solve for the minimum speed at B. Once the minimum speed is found, we can determine hmin using the work-energy theorem.)

Explanation / Answer

a)

Ea = m*g*h

Eb = 0.5*m*v^2 + m*g*2R

from energy conservation

Eb = Ea

m * g* 2R + 0.5*m*v^2 = m*g*h

v = sqrt(2*g*(h-2R))

v = 10.84 m/s

#b)

at B

N - m*g = -m*v^2/R

N = m*g - mv^2/R

N = (2.5*9.8) - ((2.5*10.84*10.84)/(1.5))

N = -171 .34 N

c) to complete the circle N shuld be minimum i.e N = 0

v1 = sqrt(gR)

sqrt(2*g*(h min-2R)) = sqrtgR

2*g*( h min -2R) = g*R

2*(h min - 3) = 1.5

h min = 3.75 m

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