Question

A 1.00 kg block is at rest at the top of a smooth hemispherical surface of radius R = 2.00 m. It is attached to anunstretched cord with stiffness of 5.00 N/m as shown on the left. It is given a slight nudge to the right, and it slides along the surface. When the block has moved to the position shown on the right what is:
a) the stretch in the cord as an exact multiple of R?
b) the speed of the block?

c) the magnitude of the normal force acting on the block by the surface?

d) what is the tangential acceleration at this instant?

(g = 9.81 m/s2.)

unstretched elastic cord smooth surface no R

Answer 1

P unstretched elastic cord smooth surface Rmg co VaR R R sin 60 60°C mg" Á RCR BAR CE R cos60

(1) stretch in the chord is obtained by calculating the elongated length AD as shown in middle figure.

AD = [ AE² + DE² ]^1/2 = [ (R + Rco60)² + ( Rsin60 )² ] = $\sqrt{3}$ R = 3.463 m

stretch in the chord $\Delta$ R = 3.463 - 2 = 1.463 m

(b) Speed of the block is obtained by calculating kinetic energy at point D on the hemisphere.

Kinetic energy at point D is difference between potential energy at P and D.

(1/2)mv² = mg(R- R sin60) = ( 1 $\times$ 9.8 $\times$ 2 $\times$ 0.134 ) J = 2.626 J

speed of block v = [ (2 $\times$ 2.626 ) ]^1/2$\approx$ 2.3 m/s

(c) magnitude of normal force

When block is at point D, there are two forces, i.e. (1) weight mg acting vertically, (2) elastic force ( k $\Delta$ R ) acting along the chord as shown in right most figure

weight mg = 9.8 N ; elastic force = k $\Delta$ R = 5 $\times$ 1.463 = 7.315 N

angle $\theta$ between these two forces :- tan$\theta$ = AE/DE = ( R + R cos60)/R sin60 = $\sqrt{3}$ or $\theta$ = 60^o

Resultant of two forces is obtained from law of parallelogram

1200 mg →

Resultant force , R_f = [ m² g² + ( kx )² - 2 (mg) (kx) cos120 ]^1/2

by substituting values , m = 1 kg, k = 5 N/m, x = $\Delta$ R = 1.463 m in above relation we get,

Resultant force R_f = 14.87 N

If resultant force R_f makes angle $\alpha$ with vertical force mg, then $\alpha$ is obtained from

kx = [ R_f² + (mg)² - 2 (mg) (R_f) cos$\alpha$ ]^1/2

we get $\alpha$ = 25^o

since weight mg acting vertically makes 30^o with normal ( refer middle figure ) at point D,

resultant force makes 25^o with vertical force mg, angle between resultant force and normal = 30-25 = 5^o

magnitude of normal force = R_f cos(30-$\alpha$) = 14.87 $\times$ cos5 = 14.81 N

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Tangential acceleration = ( R_f sin(30-$\alpha$) /m) = ( 14.87 $\times$ sin5) ) = 1.296 m/s²