The small mass m sliding without friction along the looped track shown in the fi

Question

The small mass m sliding without friction along the looped track shown in the figure(Figure 1) is to remain on the track at all times, even at the very top of the loop of radius r.

A

Determine the minimum release height h.

Express your answer in terms of r.

B.

If the actual release height is 2 h, calculate the normal force exerted by the track at the bottom of the loop.

C

If the actual release height is 4 h, calculate the normal force exerted by the track at the top of the loop.

D

If the actual release height is 2 h, calculate the normal force exerted by the track after the block exits the loop onto the flat section.

Explanation / Answer

A)

at the top ,

- N - mg = - mv^2 / r

N = 0 in extreme case ;

v = sqrt ( g r )

difference of PE = differecne of KE ;

m g ( h - 2r ) = 0.5 m v^2

from 1 ,

g ( h - 2 r ) = 0.5 ( gr )

the minimum release height h = 2.5 r

B)

at the bottom

N - mg = mv^2 / r ;

N = m ( g + v^2 / r ) ;

v = sqrt ( 2 g * 2 h ) = sqrt( 4 g h )

N = m [ g + ( 4 gh / r ) ]   

N = mg [ 1 + ( 4 * 2.5 ) ]

= mg [ 1 + 4 *2.5 ]

N=11mg

the normal force exerted by the track at the bottom of the loop. N =11 mg

C)

release height = 4h

at top

- N - mg = - m v ^2 / r ;

N = m ( v^2 / r - g ) ;

v = sqrt [ 2 g ( h - 2r ) ]

height = 4 h = 4 * 2.5 r =10 r

v = sqrt [ 2 g * 8 r ]

= sqrt(16 gr )

N=m((16gr/r)-(g))=15mg

the normal force exerted by the track at the top of the loop =15mg

D)

no vertical forces act on the block except mg at the flat section. N=mg

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