1.A small 2.0 kg mass is attached to a Hooke\'s Law spring with spring constant

Question

1.A small 2.0 kg mass is attached to a Hooke's Law spring with spring constant 64 N/m and an equilibrium length of 0.50 m. The spring and the mass lie in a frictionless channel as shown in the figure (the mass and channel are not drawn to scale) with the other end of the spring attached to a pivot at the axis of rotation of a horizontal spinning disk.   The disk rotates at a constant 4.0 rad/s. The mass can only be moved in the radial direction. At what distance from the pivot can the small mass (you can treat it as a point) be positioned so that it only undergoes uniform circular motion?

2.Two masses m1 and m2, with m1 = 3 m2, undergo a head-on elastic collision. If the particles were approaching with speed v before the collision, with what speed are they moving apart after collision?

A. v/3

B. v

C. 3v

D.3v/4

Explanation / Answer

For a body to perform a uniform circular motion, the centrifugal force it experiences should equal the centripetal force that acts on it.

Centrifugal force = m*r*w^2

where m is mass of body

r is radius of circle

w is angular velocity.

So, C.F force = 2*r*4^2

= 32r

Centripetal force = Force due to elongation in spring.

Elongation in spring = total length - equilibrium length

= r-5.0

Force by spring = 1/2 * k*x^2

= 1/2 *k*(r-5.0)^2

= 32*(0.5-r)^2

C.P force = C.F force

So, 32r = 32*(0.5-r)^2

r^2 + 0.25 -r = r

r = 1.86 m

Note: Not more than one question can be asked in a single post. Please post your next question seperately. Thank you

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.