The net external force acting on the disk when its centre of mass is at displacement x with respect to its equilibrium position is

k x

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 k x

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- \frac{2 k x}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- \frac{4 k x}{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $- \frac{4 k x}{3}$
$2 k x - f = m a$

Since we know that $Γ = I α$ For disk, $I = \frac{M R^{2}}{2}$

Also, $a = R α$

Hence, $m a = \frac{4 k x}{3}$

Suggest Corrections

Similar questions

Comprehension Type: A uniform thin cylindrical disk of mass M ad radius R is attached to two identical mass less springs of springs constant k which are fixed to the wall as show in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance d from its center. The axle is mass less and both the springs and the axle are in horizontal plane. The un-stretched length of each springs is L. The disk is initially at its equilibrium position with its center of mass (CM) at a distance L from the wall. The disk rolls without slipping with velocity $\to (V)_{0} = V_{0}^i$ The coefficient of friction is $μ$ .

The net external force acting on the disk when its centre of mass is at displacement x with respect to its equilibrium position is

Q. A uniform thin cylindrical disk of mass M and radius R is attached to two identical massless springs of spring constant k which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance d from its centre. The axle is massless and both the springs and the axle are in horizontal plane. The un-stretched length of each spring is L. The disk is initially at its equilibrium position with its centre of mass (CM) at a distance L from the wall. The disk rolls without slipping with velocity

_{0} = V_{0}^i

. The coefficient of friction is

μ

.
The net external force acting on the disk when its centre of mass is at displacement x with respect to its equilibrium position is?

When external forces acting on a body are zero, then its centre of mass

An object moves with a uniform velocity of 5 m/s when the net force acting on it is___.

Q. If the displacement

x

of a particle of mass

m

is directly proportional to time, net force acting on it is _____________.

Join BYJU'S Learning Program

The net external force acting on the disk when its centre of mass is at displacement x with respect to its equilibrium position is

The correct option is D −4kx32kx−f=maSince we know that Γ=Iα For disk, I=MR22Also, a=RαHence, ma=4kx3

The correct option is D $- \frac{4 k x}{3}$
$2 k x - f = m a$

Since we know that $Γ = I α$ For disk, $I = \frac{M R^{2}}{2}$

Also, $a = R α$

Hence, $m a = \frac{4 k x}{3}$